Overview
This unit on Current Electricity covers electric current, resistance, resistivity, Ohm’s law and their microscopic basis, how resistors combine in circuits, and methods to analyse circuits such as Kirchhoff’s laws. It treats practical sources of emf and their internal resistance, power and energy in circuits, heating effects and how real devices deviate from ideal behaviour. Measurement techniques — potentiometer, meter-bridge, Wheatstone bridge, ohmmeters and graphical methods — are explained along with experimental precautions, errors and data analysis. The unit also introduces drift velocity, mobility and a simple microscopic model linking material properties to observed conductivity. Understanding these topics prepares students to solve circuit problems, perform laboratory measurements accurately and reason about everyday devices like batteries, heaters and lamps. Emphasis is placed on linking formulas to physical meaning, interpreting graphs and following safe laboratory practice. Mastery of this unit is important for board exams and lays groundwork for electronics, power systems and experimental physics at higher levels.
Learning Objectives
- Describe electric current as the rate of flow of charge and relate it to microscopic drift velocity of carriers.
- Explain and apply Ohm’s law and distinguish ohmic from non-ohmic behaviour.
- Calculate resistance from resistivity and geometry and explain how temperature and impurities affect resistivity.
- Combine resistances in series and parallel and analyse mixed networks to find currents and voltages.
- Apply Kirchhoff’s junction and loop rules to solve circuits that cannot be reduced by simple series-parallel combinations.
- Determine emf and internal resistance of a cell experimentally and by graphical methods.
- Compute electrical power and energy in circuits and relate them to heating effects and device ratings.
- Use potentiometer and bridge methods for precise measurement of emf and resistance while accounting for errors and precautions.
Topics in this chapter
15 topics · tap a topic title to jump straight to it.
Electric Current and Charge Flow
Definition and macroscopic view. Electric current I is the rate at which electric charge Q passes through a cross-section: I = dQ/dt. In everyday circuits we speak of current as a scalar quantity measured in amperes (A). Conventional current direction is taken from higher to lower potential, though in metallic conductors the mobile charge carriers are electrons moving opposite to that direction. Drawing a clear arrow for assumed current direction helps in solving circuit problems; a negative result indicates the actual current flows opposite to the assumed arrow.
Microscopic picture and drift velocity. On a microscopic scale charges have large random thermal velocities but no net flow without an external field. When an electric field E is applied, carriers gain a small average velocity called the drift velocity v_d. The drift is superimposed on random motion so individual paths are zigzag due to frequent collisions; nonetheless the collective effect is a measurable current. For a conductor with number density n of charge carriers, cross-sectional area A and charge e per carrier, current is I = n A e v_d. This formula is useful for estimating v_d from measured currents and shows why tiny drift speeds still produce large currents when n and A are large.
Current density and local relations. Current density J is a vector field defined as current per unit area: J = I/A for uniform conductors and generally J = n e v_d. In many materials J and E are locally related by J = σ E where σ is conductivity; this local form of Ohm’s law connects macroscopic circuit behaviour to fields inside the conductor. The continuity equation ∂ρ/∂t + ∇·J = 0 expresses charge conservation, where ρ is local charge density; in steady-state DC circuits ∇·J = 0, meaning currents into and out of any node balance.
Practical measurement and safety notes. Ammeters measure current and must be connected in series; they are designed to have very low internal resistance to avoid changing circuit currents. Wires and devices have current ratings; excessive current causes heating and potential hazards. When designing circuits, choose appropriate wire gauge and fuse ratings. In laboratory work record directions and magnitudes carefully, and be aware that although signals propagate rapidly along the circuit, individual electrons move slowly. Understanding both macroscopic and microscopic views helps bridge practical circuit analysis and physical intuition about materials and devices.
- Compute electron drift velocity in a copper wire carrying 2 A given n and A.
- Find number of electrons passing a cross-section per second for a 3 A current.
- Use the continuity idea to explain equal currents in series elements in steady state.
- I = dQ/dt
- I = n A e v_d
- J = I/A
- J = σ E
Ohm's Law, Resistance and Resistivity
Ohm’s law — empirical statement. For many materials at constant temperature the potential difference V across a conductor is proportional to the current I through it: V = IR. Materials that obey this linear relation are called ohmic; their V–I graph is a straight line through the origin. The law is local in nature when written as J = σ E, linking current density to electric field. However, Ohm’s law is not a fundamental law of nature; it is an empirical observation valid for many conductors within certain ranges.
Resistance as geometry and material property. The resistance R of a uniform cylindrical conductor of length l and cross-sectional area A is given by R = ρ l / A where ρ is resistivity intrinsic to the material. Resistivity depends on electronic structure and scattering processes. Conductivity σ = 1/ρ provides a measure of how well a material conducts. For design purposes choose materials and dimensions to meet required resistance values: long thin wires have high resistance, short thick conductors have low resistance.
Microscopic origin. In the classical Drude picture, electrons accelerate under the electric field between collisions with lattice ions; the average relaxation time τ between collisions leads to drift velocity v_d = e E τ / m. From this emerges conductivity σ = n e^2 τ / m, connecting macroscopic conductivity to microscopic parameters: carrier density n, charge e, mass m and scattering time τ. Thus impurities, defects and phonons (lattice vibrations) reduce τ and increase resistivity.
Temperature dependence and non-idealities. For many metals resistivity rises with temperature approximately linearly over moderate ranges: ρ(T) = ρ0[1 + α (T − T0)], where α is the temperature coefficient. Semiconductors generally show decreasing resistivity with temperature because carrier concentration increases. In real devices resistance can change with applied voltage and heating; filament lamps show marked nonlinearity as temperature changes with current. Always state temperature when giving resistivity values and be cautious using Ohm’s law for non-ohmic materials.
Measurement considerations. For accurate resistivity measurement use four-probe methods for low resistances to avoid lead and contact errors. Calibrate instruments and control temperature. Understanding R = ρ l / A helps in material selection for wires and resistors and in interpreting how geometry affects circuit behaviour.
- Calculate resistance of a copper wire of given length and area using R = ρl/A.
- Compute change in resistance of a resistor heated by ΔT using R(T) = R0[1 + αΔT].
- Explain why filament bulbs show curved V–I characteristics as they heat up.
- V = IR
- R = ρ l / A
- σ = 1/ρ
- ρ(T) = ρ0[1 + α (T - T0)]
Series and Parallel Resistances and Mixed Networks
Series connection — rules and consequences. When resistors are connected end to end in a single path they are in series and the same current flows through each. The equivalent resistance is the sum: R_eq = R1 + R2 + ... + Rn. Voltage from the source divides among resistors according to their resistances: V_i = I R_i. Series combinations are used where current limiting is required. Power dissipated in each resistor is P_i = I^2 R_i, so larger resistances dissipate more of the supply voltage as heat for a given current.
Parallel connection — rules and consequences. Resistors connected across the same two nodes are in parallel; they share the same voltage but carry different currents. The equivalent resistance satisfies 1/R_eq = 1/R1 + 1/R2 + ... + 1/Rn. Parallel networks reduce total resistance and increase total current draw from the source. Currents split inversely proportional to resistances: I_i = V / R_i. Parallel wiring is common in household circuits so appliances receive full supply voltage independently.
Mixed networks — stepwise reduction and node analysis. Real circuits often combine series and parallel parts. Solve by identifying clearly series pairs and parallel groups, replace them with their equivalents, and repeat until the network is simplified. Keep a record of intermediate equivalents so you can back-solve to find individual branch currents or voltages. When elements are neither purely series nor purely parallel (e.g. bridge networks), series-parallel reduction fails and one must use Kirchhoff’s laws.
Use of conductance for parallel sums. It is often convenient to work with conductance G = 1/R; for parallel branches conductances add directly: G_eq = G1 + G2 + ... . This simplifies mental calculations and helps check answers quickly. Also consider limiting cases: if one branch resistance → ∞ it is effectively open; if one branch → 0 it shorts the nodes and dominates current flow.
Practical considerations and safety. Always ensure component power ratings are not exceeded: compute P = V^2 / R or P = I^2 R for each resistor and select suitable wattage. For series strings powering multiple devices, a failure in one opens the circuit; in parallel this does not happen. Use these ideas to design circuits safely and to interpret how voltage and current distribute in complex networks.
- Find equivalent resistance of 2 Ω, 3 Ω and 6 Ω in series and in parallel.
- Simplify a circuit with two resistors in series connected in parallel with a third resistor and find total current for a given battery voltage.
- Compute voltage drops after stepwise reduction and verify power sum.
- R_eq(series) = Σ R_i
- 1 / R_eq(parallel) = Σ (1 / R_i)
Electromotive Force, Internal Resistance and Terminal Voltage
Ideal emf versus real source. An ideal emf ε is the work done per unit charge by non-electrostatic forces inside a device when no current flows; it equals the open-circuit voltage. A real cell or battery has internal resistance r arising from chemical processes, electrode resistance and ionic resistance in the electrolyte. When delivering current I the terminal voltage V is reduced: V = ε - I r. This simple series model (ε in series with r) is a powerful way to analyse battery behaviour.
Circuit consequences and short-circuit current. If an external resistance R is connected across the cell, current I = ε / (R + r) flows. A short circuit (R = 0) gives I_sc = ε / r, a potentially large damaging current limited only by internal resistance; this illustrates why shorting batteries is dangerous. When charging a cell with current forced into it, terminal voltage becomes V = ε + I r (sign convention reversed), and over-voltage can cause heating and electrochemical damage.
Measurement techniques. Measure ε and r by varying external load and recording V and I. Plot V against I to get a straight line: intercept at I = 0 gives ε and slope equals -r. Potentiometer methods allow direct measurement of ε without drawing current and so avoid internal resistance effects, giving more accurate emf values. Combining potentiometer-determined ε with loaded terminal voltage readings yields precise r values using V = ε - I r.
Energy and efficiency. Power delivered to load is P_R = I^2 R while power dissipated inside the source is P_r = I^2 r. Maximum power transfer to the load occurs when R = r but half the power is wasted inside the source — an inefficient operating point for power supplies. Practical cells aim for r much smaller than expected loads to keep heating low and efficiency high.
Practical notes and safety. Avoid short circuits and overcurrent charging. For battery packs, match cells to avoid imbalanced currents and possible overheating. In exam problems explicitly state sign conventions, model the source as ε with series r, and check units and magnitudes for plausibility.
- Given ε = 12 V and r = 0.5 Ω connected to R = 5 Ω, compute I and V_term.
- From V vs I data, determine ε and r by linear fitting.
- V_term = ε - I r
- I = ε / (R + r)
- I_sc = ε / r
Kirchhoff’s Laws and Circuit Analysis
Fundamental statements. Kirchhoff’s laws are essential tools for analysing circuits with multiple branches and sources. The junction rule (first law) states that the algebraic sum of currents entering a junction equals the sum leaving it — this is conservation of charge. The loop rule (second law) states that the algebraic sum of potential differences around any closed loop is zero — conservation of energy. These laws are universally valid for circuits in steady state where magnetic flux linking loops is constant.
Systematic method to apply laws. To apply Kirchhoff’s rules reliably follow steps: (1) Draw the circuit with clear labeling of nodes, branches and components. (2) Assign an arbitrary current direction in each branch; choose loop directions for applying loop equations. (3) Write junction equations for independent nodes. (4) Write independent loop equations ensuring you do not write redundant ones. (5) Use consistent sign conventions: when traversing a resistor in direction of its assumed current, subtract IR; when traversing opposite add IR. For emf, traversing from negative to positive terminal adds +ε, and opposite gives -ε. (6) Solve the resulting linear equations by substitution, elimination or matrix methods.
Interpreting results and checking consistency. If a calculated current is negative the assumed direction is opposite to real direction — this is acceptable and conveys physical information. Check power balance: power supplied by emf sources should equal power dissipated in resistors (including internal resistances). Also verify limiting cases: if a resistor value is very large or small does the result approach expected behaviour. Consistency checks help identify algebra or sign mistakes before finalising your answer.
Examples where Kirchhoff is necessary. Bridge circuits, multiple sources, and networks that are not reducible by series-parallel simplifications require Kirchhoff’s approach. For exam solutions show clear steps: chosen loop directions, equations, algebraic solution and final numeric answers with units. For circuits with time-varying magnetic flux the loop rule must be modified to include induced emf; this is beyond DC steady-state scope but worth noting.
Practical tips. Choose loop directions conveniently to simplify signs, reduce number of unknowns where possible, and use symmetry when present. Label intermediate results to avoid confusion when back-substituting to find branch voltages or powers. Good presentation and logical steps earn marks in board exams.
- Solve a two-loop circuit with two emfs and three resistors using loop and junction equations.
- Analyse an unbalanced Wheatstone bridge using Kirchhoff’s rules to find currents.
- Σ I_in = Σ I_out (junction rule)
- Σ ΔV_around loop = 0 (loop rule)
Power, Energy and Heating in Electric Circuits
Definitions and basic formulas. Electric power P is the rate at which electrical energy is converted into other forms, defined as P = VI for a device across potential difference V carrying current I. Using Ohm’s law this becomes P = I^2 R or P = V^2 / R for resistive elements. Energy consumed over time t is E = P t. In practical billing units energy is measured in kilowatt-hours (kWh): 1 kWh = 3.6 × 10^6 J.
Joule heating and mechanism. Joule’s law quantifies heat produced by current: H = I^2 R t. Microscopically, moving charge carriers lose kinetic energy through collisions with the lattice, transferring energy to the material as heat. Heat dissipation causes temperature rise depending on heat capacity and heat loss; continuous high power can damage components unless designed for or cooled adequately.
Applications and design choices. Heaters, toasters and electric irons rely on controlled Joule heating; heating elements are made from materials with appropriate resistivity and high melting points. Filament lamps convert electrical energy into light and heat; filament resistance rises with temperature, producing non-linear characteristics. For power transmission, losses in lines are P_loss = I^2 R_line; to minimise losses power is transported at high voltages to reduce current for the same power, then stepped down by transformers for local usage.
Internal resistance and efficiency. A source with emf ε and internal resistance r supplies load R with current I = ε/(R + r). Power delivered to the load is P_R = I^2 R and power wasted inside the source is P_r = I^2 r. Maximum power transfer to the load occurs when R = r, giving P_max = ε^2 / (4 r), but only 50% of generated power is delivered to the load in that case — not efficient for power systems, which prefer r ≪ R.
Practical calculations and safety. Always compute expected power dissipation and ensure components and wiring have adequate power ratings and cooling. Fuses and circuit breakers protect against excessive currents by interrupting circuits before dangerous heating occurs. In exam answers state formula used, perform unit-consistent substitution and report final answers with units and appropriate significant figures.
- Calculate power and energy consumed by a 100 Ω heater at 230 V for 2 hours in kWh.
- Compute power loss in a transmission line carrying 200 A with resistance 0.05 Ω.
- P = VI
- P = I^2 R
- P = V^2 / R
- E = P t
- H = I^2 R t
- P_max = ε^2 / (4 r) when R = r
Heating Effect, Joule’s Law and Practical Implications
Statement and derivation of Joule’s law. Joule’s law states that the heat H produced in a resistor of resistance R carrying current I for time t is H = I^2 R t. The result follows by integrating instantaneous power P = I^2 R over time. This expression links measurable electrical quantities to thermal effects and is the basis for resistive heating devices.
Microscopic origin and material considerations. When charge carriers move under an electric field they collide with lattice ions and imperfections, losing kinetic energy that is transferred to the lattice as vibrational energy (heat). Materials chosen for heating elements (nichrome, Kanthal, tungsten) have appropriate resistivity, melting point and oxidation resistance so they can operate at high temperatures without rapid degradation. The temperature coefficient of resistance influences how heating elements change resistance with temperature, affecting current and power behavior during operation.
Applications and design trade-offs. Electric heaters, water heaters and toasters exploit Joule heating; design requires selecting resistance so that at supply voltage desired power is dissipated without exceeding material temperature limits. Filament lamps rely on high-temperature filaments to emit visible light; however, filament resistance rises with temperature, causing non-linear V–I behaviour and significant power loss as heat. Safety devices like fuses protect against sustained overcurrent by melting at specified current thresholds, interrupting dangerous heating before damage occurs.
Measurement and experimental methods. Heat produced can be measured calorimetrically by immersing the resistor in water and measuring temperature rise; compare electrical energy input E_elec = V I t with thermal energy gained mcΔT adjusting for losses. Sources of experimental error include heat loss to surroundings, non-uniform heating and inaccurate current or voltage readings. Reduce errors by insulating the calorimeter and taking short-duration runs to limit losses.
Energy efficiency and power systems. In power distribution, I^2R losses motivate transmitting power at high voltage to reduce current, thereby lowering line losses. For safety and reliability always ensure wiring and components have suitable current and power ratings, leave thermal margins and include overcurrent protection. In answers show formula used, substitute values with units and provide final value with correct significant figures.
- Compute heat produced in a 10 Ω resistor carrying 2 A for 5 minutes using H = I^2 R t.
- Explain why in household wiring using thicker conductors reduces heating for the same current.
- H = I^2 R t
- P = I^2 R = V^2 / R = VI
Measurement of Resistance: Ohmmeter, Meter Bridge and Wheatstone Bridge
Direct methods and their limitations. Ohmmeters apply an internal voltage and measure current to compute resistance; they are convenient for quick checks but accuracy is limited by internal calibration, temperature effects and the influence of parallel circuit elements if measured in-circuit. For very low resistances contact and lead resistances dominate, while for very high resistances leakage currents and insulation resistance can give large errors.
Meter bridge — practical null method. The meter bridge uses a long uniform wire mounted on a scale, a known standard resistor and the unknown resistor forming two arms, with a galvanometer and jockey used to find the balance point. At balance, no current flows through the galvanometer and R_x / R_std = l_1 / l_2 where l_1 and l_2 are lengths on the bridge wire. Because the galvanometer sees zero current at balance, measurement is not affected by galvanometer internal resistance, leading to good precision for moderate resistance values. Ensure wire uniformity, good contact and stable temperature for accurate results.
Wheatstone bridge — sensitive and precise. A true Wheatstone bridge has four resistances in a diamond configuration with a galvanometer between two opposite nodes and a supply across the other nodes. Balance occurs when R1/R2 = R3/R4; at this point the galvanometer shows zero. The bridge can detect minute resistance changes and is used in instrumentation such as strain gauges. For low-resistance measurements a Kelvin (four-terminal) bridge separates current and potential leads to eliminate lead resistance errors.
Practical tips and error sources. Minimise contact resistance by cleaning contacts, use a sensitive galvanometer to detect null accurately, and repeat balance measurements to average out random errors. Temperature affects wire resistance; reduce drift by performing measurements quickly or in a temperature-controlled environment. For low-resistance work use Kelvin connections; for high-resistance measurements use guarding to reduce leakage paths. Report measured values with an estimate of uncertainty accounting for instrument precision and repeatability.
- Using a meter bridge with total wire length 1 m and balance point at 40 cm, find unknown resistance when standard is 10 Ω.
- Explain how a Kelvin connection removes lead resistance error when measuring a 0.01 Ω shunt.
- R_x / R_std = l_1 / l_2 (meter bridge balance)
- R1 / R2 = R3 / R4 (Wheatstone balance)
Potentiometer: Principle, Measurement of Emf and Internal Resistance
Principle of null method. A potentiometer compares an unknown emf to the potential difference across a length of a long uniform wire carrying a steady current from a driving cell. By sliding a jockey along the wire until the galvanometer reads zero, the potential drop between the contact points equals the emf under test. Because no current flows through the test cell at the null point, the measurement is free from loading effects and provides high accuracy.
Measuring emf accurately. Set up the potentiometer with a stable driving cell and a rheostat to control the current. Use a standard cell of known emf to determine the potential gradient k (voltage per unit length) or calibrate directly. Place the test cell in series with the galvanometer and connect across a portion of the potentiometer wire; vary jockey position until the galvanometer gives null and measure the balancing length l. The unknown emf ε = k l. Using a potentiometer avoids errors due to voltmeter internal resistance and provides more precise values for small emfs.
Determining internal resistance using potentiometer. First determine open-circuit emf ε by null method. Then connect a known external resistor across the test cell to draw current and measure the new terminal voltage V by finding the balancing length l'. The internal resistance r can be obtained from V = ε - I r where I is current supplied (measured by an ammeter). A more accurate approach repeats measurements for several loads and plots V against I; the intercept gives ε and the slope equals -r, reducing random error.
Precautions and accuracy. Keep the driving current stable, avoid heating the potentiometer wire as it changes its resistance and hence potential gradient, ensure good electrical contacts and use a sensitive galvanometer for precise null detection. Reverse connections properly to check polarity. For exam answers outline apparatus, procedure, equations used and typical precautions succinctly.
- Given potential gradient 0.01 V/cm, find balancing length for a cell of emf 1.5 V.
- Outline steps to obtain internal resistance of a cell using potentiometer readings under load and open-circuit.
- ε = k l
- V = k l' (terminal voltage under load)
- V = ε - I r
Drift Velocity, Mobility and Microscopic Model of Conduction
Drift velocity and mobility. Drift velocity v_d is the small average velocity acquired by charge carriers in a conductor under an applied electric field. It is given by v_d = μ E where μ is the mobility and E the electric field. Mobility quantifies how readily carriers respond to the field and depends on scattering rates and temperature. Using I = n A e v_d connects microscopic carrier motion to measurable current in a conductor of cross-sectional area A and carrier density n.
Drude model and conductivity. The classical Drude model describes conduction electrons as particles undergoing random collisions with lattice ions; between collisions they accelerate under the field and on average lose momentum at collisions. The mean free time τ between collisions yields v_d = e E τ / m and conductivity σ = n e^2 τ / m. This links macroscopic conductivity to microscopic parameters: carrier density n, charge e, mass m and relaxation time τ. Mobility μ = e τ / m and σ = n e μ.
Orders of magnitude and signal propagation. Typical drift velocities are very small (10^-4 to 10^-3 m s^-1) for ordinary currents, yet signal propagation is near light speed because the applied field propagates rapidly along the conductor and sets electrons into motion almost simultaneously. Thus the electrical signal appears to travel fast while individual carriers move slowly. In semiconductors both electrons and holes contribute with different mobilities; doping modifies n and therefore conductivity significantly.
Limitations and practical use. The Drude model is classical and cannot explain quantum phenomena like electron degeneracy effects, but it provides useful estimates for τ, μ and mean free path from measured conductivity and carrier density. These estimates help in materials selection and understanding temperature and impurity effects on resistivity. In lab problems, given σ and n one can estimate τ and μ to relate experimental measurements to microscopic scattering processes.
- Given copper carrier density and conductivity, estimate mean free time τ and mobility μ.
- Calculate drift velocity in a wire carrying 5 A with given n and A.
- v_d = μ E
- μ = e τ / m
- I = n A e v_d
- σ = n e^2 τ / m
Non-Ohmic Conductors, Dynamic Resistance and Temperature Effects
What is non-ohmic behaviour? A non-ohmic device does not show a constant ratio V/I over its operating range; its V–I graph is nonlinear. Common examples include diodes, Zener diodes, gas discharge tubes, filament lamps and thermistors. For such devices the instantaneous slope at an operating point defines dynamic or differential resistance r_d = dV/dI. For small perturbations around an operating point, the device may be approximated by this r_d to analyse circuits with linear methods.
Filament lamps and temperature dependence. Filament bulbs are classic examples: when cold the filament has low resistance and draws a large inrush current. As current flows the filament heats up, increasing resistivity sharply; this raises resistance and reduces steady-state current. The resulting V–I curve is concave, not a straight line. The temperature coefficient of resistance α for the filament material is large so even modest current changes can alter resistance significantly.
Thermistors and sensors. Thermistors are resistors whose resistance changes strongly with temperature. NTC (negative temperature coefficient) thermistors decrease resistance as temperature rises and are used for temperature sensing and inrush current limiting. PTC (positive temperature coefficient) thermistors increase resistance with temperature and find uses in self-regulating heaters and overcurrent protection. Their strong temperature dependence makes them useful as sensors but requires careful calibration and compensation in circuits.
Semiconductor devices and nonlinearity. Diodes have exponential I–V characteristics in forward bias and large resistance in reverse bias until breakdown. These behaviours are exploited in rectification and switching. For circuit design, engineers often linearise nonlinear elements about an operating (bias) point and use the dynamic resistance for small-signal analysis; this simplifies AC analysis and amplifier design.
Experimental and practical notes. When measuring resistance of non-ohmic devices specify the applied voltage or current because the measured R depends on operating point. Avoid heating effects by using low measurement currents or short measurement durations, and note that temperature drift can change readings. In exam answers define dynamic resistance and explain device behaviour with respect to temperature and operating conditions.
- Sketch V–I characteristic of a filament lamp and explain why it is non-linear.
- Explain why an NTC thermistor is used inrush current limiter in power supplies.
- r_d = dV / dI
- ρ(T) = ρ0[1 + α (T - T0)]
Capacitors in DC Circuits: Transients and Steady State
Role of capacitors in DC circuits. Capacitors store charge Q = C V and influence circuit behaviour during changes. When a DC source is suddenly applied, capacitors allow transient currents while charging, but once fully charged they behave as open circuits and block steady DC current. This makes them useful for coupling and blocking DC while passing AC components in mixed-signal circuits.
Transient charging and discharging. For a simple series RC circuit with supply ε and resistor R, the capacitor charge follows q(t) = C ε (1 - e^{-t/RC}) and the current i(t) = (ε/R) e^{-t/RC}. The time constant τ = RC governs the rate: after a time of about 3τ the capacitor is ~95% charged, and after 5τ it is effectively fully charged for practical purposes. During discharge through R the charge decays exponentially: q(t) = Q_0 e^{-t/RC} and i(t) = (Q_0/RC) e^{-t/RC}.
Practical implications for measurements and circuits. When taking DC measurements in circuits containing capacitors allow adequate time for transients to settle; otherwise measured currents or voltages may reflect transient values, leading to incorrect conclusions. In timing circuits RC networks produce predictable delays used in timers and pulse shaping. In power supplies capacitors smooth ripple by charging when voltage is high and discharging when it falls, reducing AC variation on DC rails.
Safety and handling. Large capacitors can store significant energy and remain charged after power is removed; always discharge capacitors through a resistor before handling to avoid shocks and component damage. For experimental problems explicitly state whether you analyse transient or steady-state behaviour. In steady-state DC problems treat fully charged capacitors as open circuits; redraw the circuit excluding capacitor branches that become open to simplify analysis.
Further notes. Non-ideal capacitors have leakage resistance and equivalent series resistance (ESR) affecting long-term behaviour and heating. For exam answers include relevant formulas, clearly state assumptions (ideal capacitor, initial uncharged), and show stepwise calculation of q(t), i(t) and voltages with units.
- For R = 1 MΩ and C = 1 μF calculate time constant τ and time to reach 95% charge.
- In a DC circuit containing a capacitor and resistors, describe the circuit after a long time since switching.
- \[q(t) = C ε (1 - e^{-t/RC})\]
- \[i(t) = (ε / R) e^{-t/RC}\]
- τ = R C
Measurement of Emf and Internal Resistance by Graphical Method
Linear relation for terminal voltage. A source with emf ε and internal resistance r has terminal voltage V under load I given by V = ε - I r. This linear relation suggests a simple graphical technique: plot measured terminal voltages V (y-axis) against corresponding currents I (x-axis). The straight line’s intercept at I = 0 equals ε and its slope is -r.
Experimental procedure and data handling. Use a variable external resistor or rheostat to obtain several (V, I) pairs: connect an ammeter in series and voltmeter across the cell, vary load and record readings. Choose ranges so meter resolutions are suitable and avoid heavy loads that heat the cell. Plot V vs I with properly scaled axes and draw a best-fit straight line (least-squares or careful eye-fit). Read intercept for ε and compute r = -slope. Multiple points reduce random error and reveal systematic deviations if the plot is not linear (e.g. due to heating).
Combining potentiometer for accuracy. Measure ε precisely using a potentiometer (null method) to avoid loading errors. Then measure V under several loads and determine r from V = ε - I r using r = (ε - V)/I or via slope from a V vs I plot. Potentiometer-determined ε improves accuracy of r since it reduces uncertainty in intercept.
Uncertainty estimation and precautions. Include instrument uncertainties and scatter in points when reporting ε and r. Allow cell to rest between heavy loads, avoid heating wires and contacts, and ensure meters are suitable: high internal resistance voltmeter and low resistance ammeter minimize measurement disturbance. In exam answers sketch the graph showing intercept and slope, state equations used and present numeric results with units and uncertainty where appropriate.
- Given a table of I and V data, draw V vs I and find ε and r from intercept and slope.
- Explain why potentiometer measurement of ε is more accurate than direct voltmeter reading.
- V = ε - I r
- Slope = -r, Intercept = ε
Conductivity of Electrolytes and Ohm’s Law in Solutions
Ionic conduction and carriers. In electrolytic solutions current is carried by ions: positive cations move toward the negative electrode and negative anions move toward the positive electrode. Mobility of ions depends on their size, charge, solvent viscosity and temperature. Unlike metallic conduction where electrons carry charge, electrolytic conduction often involves electrode reactions and concentration changes near electrodes that influence observed behaviour.
Conductance, conductivity and cell constant. Conductance G (siemens) is the reciprocal of resistance R. For a conductivity cell, measured conductance G relates to the sample conductivity κ by κ = K G where K is the cell constant determined by electrode geometry (approximately distance between electrodes divided by electrode area). Calibration with standard solutions is required to find K accurately. Report conductivity at specified temperature since κ depends strongly on temperature.
Limits of Ohm’s law in electrolytes. Ohm’s law (linear I–V relation) often holds approximately for dilute solutions under moderate applied potentials. At high applied voltages or with reactive electrodes, nonlinearities arise due to concentration polarization, ion depletion near electrodes and Faradaic reactions. Electrode polarization in DC measurements leads to time-dependent changes; using AC techniques reduces polarization effects and yields more reliable conductance measurements.
Practical applications and precautions. Conductivity measurement is used to assess water purity, monitor industrial processes and control electroplating. For accurate results, avoid gas evolution or electrolysis at electrodes during DC measurements, use AC bridges for low conductivity samples, and control temperature. Explain electrode effects and why AC methods are preferred in exam answers. For problem solving use κ = K G and G = 1/R with attention to units.
- Given measured conductance G and cell constant K, compute conductivity κ = K G.
- Explain why distilled water conducts poorly compared to tap water and how ions determine conductivity.
- G = 1 / R
- κ = K G
Practical Circuits, Safety and Exam Revision Tips
Laboratory safety and good practice. Before building circuits, switch off power and verify component ratings for voltage, current and power. Use insulated wires, avoid exposed live contacts and mount circuits on non-conducting boards. Use current-limited power supplies or series resistors for student work and fit appropriate fuses on mains-operated circuits. For capacitors, always discharge through a resistor before handling. Ensure correct polarity for electrolytic components and check that batteries are not shorted.
Measurement techniques and instrument use. Connect ammeters in series and voltmeters in parallel. Select meter ranges so readings use a central part of the scale for better accuracy. For low-resistance measurements use four-terminal (Kelvin) connections to remove lead and contact resistance; for high-resistance values guard against leakage by insulating and using guarded techniques. In potentiometer and bridge experiments maintain stable driving current, make good contacts and repeat measurements to average out random errors.
Recording and error analysis. Record raw data with units and uncertainties where possible. State procedure, apparatus and precautions succinctly for lab questions. Identify sources of error: contact resistance, temperature variations, instrument calibration and parallax in scale readings. Where asked, estimate uncertainty propagation through calculations to show understanding of experimental reliability.
Exam strategy and problem solving. Read questions carefully and draw clear labelled diagrams. Check whether series-parallel simplification is possible; if not use Kirchhoff’s laws. Use named formulas (I = ε/(R + r), R_eq formulae, P = I^2 R) and show substitution with units. For graphical questions draw axes with labels and units, plot points accurately and indicate best-fit line; extract slope and intercept with units. Manage time by attempting high-mark problems first and keeping algebra organised to avoid sign errors.
Common pitfalls and checks. Watch sign conventions in Kirchhoff’s law, include internal resistance in battery problems, and remember that RMS and peak values differ for AC. Use limiting-case reasoning to check answers (e.g. R→0 or R→∞). Practise varied problems and laboratory exercises to build confidence for board examinations.
- List safety steps before connecting a circuit to mains in a school laboratory.
- Describe correct ammeter and voltmeter connections when measuring a resistor in a circuit.
Key Concepts
- Current (I)
- Rate of flow of electric charge, I = dQ/dt.
- Drift Velocity
- Average velocity of charge carriers due to an applied electric field.
- Current Density (J)
- Current per unit cross-sectional area, J = I/A.
- Ohm's Law
- At constant temperature, V is proportional to I and V = IR for ohmic materials.
- Resistance (R)
- Opposition to current in a conductor, related to geometry and material.
- Resistivity (ρ)
- Material property quantifying resistance per unit length and area.
- Conductivity (σ)
- Reciprocal of resistivity, σ = 1/ρ.
- Electromotive Force (ε)
- Work done per unit charge by a source in moving charge around the circuit with no current drawn.
- Internal Resistance (r)
- Intrinsic resistance within a source causing drop in terminal voltage under load.
- Power (P)
- Rate of energy transfer, P = VI = I^2 R = V^2 / R.
- Joule Heating
- Heat produced by current in a resistor, H = I^2 R t.
- Kirchhoff’s Laws
- Junction rule and loop rule used to apply charge and energy conservation to circuits.
- Potentiometer
- Null instrument to measure emf without drawing current by balancing potentials along a wire.
- Mobility (μ)
- Proportionality between drift velocity and electric field: v_d = μ E.
- RMS Value
- Equivalent DC value producing same heating effect; for sinusoid I_rms = I_0/√2.
- Wheatstone Bridge
- Bridge circuit for precise resistance measurement using balance condition R1/R2 = R3/R4.
- Continuity Equation
- Expression of charge conservation ensuring currents entering and leaving a volume balance.
- Cell Constant (K)
- Geometric factor relating conductance of a cell to sample conductivity: κ = K G.
Practice Questions
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A wire of length 2 m and cross-sectional area 1 mm^2 has resistivity 1.7 × 10^-8 Ω m. Find its resistance. / एक 2 m लंबा तार जिसकी अनुप्रस्थ छेदफल 1 mm^2 और प्रतिरोध प्रतिरोधकता 1.7 × 10^-8 Ω m है उसका प्रतिरोध ज्ञात कीजिए।
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R = ρ l / A = (1.7 × 10^-8 × 2) / (1 × 10^-6) = 0.034 Ω. / R = ρ l / A = (1.7 × 10^-8 × 2) / (1 × 10^-6) = 0.034 Ω।
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State Kirchhoff’s loop rule and apply it to a simple single-loop circuit containing one emf ε and two resistors R1 and R2 to find the current. / किर्चहॉफ के लूप नियम को लिखिए और एक सरल-लूप सर्किट जिसमें एक emf ε और दो अवरोधक R1 तथा R2 हों, में धारा ज्ञात करने हेतु लागू कीजिए।
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Loop rule: algebraic sum of potential differences around a closed loop is zero. For loop: ε - I R1 - I R2 = 0 so I = ε / (R1 + R2). / लूप नियम: बंद लूप के चारों ओर विभवांतरों का बीजगणितीय योग शून्य होता है। अतः ε - I R1 - I R2 = 0 से I = ε / (R1 + R2)।
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A cell of emf 12 V and internal resistance 1 Ω is connected to a 5 Ω resistor. Calculate terminal voltage and power delivered to resistor. / 12 V emf और आंतरिक प्रतिरोध 1 Ω वाले सेल को 5 Ω प्रतिरोध से जोड़ा गया है। टर्मिनल वोल्टेज और प्रतिरोध को दी गई शक्ति ज्ञात कीजिए।
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I = ε / (R + r) = 12 / (5 + 1) = 2 A. V_term = IR = 2 × 5 = 10 V. Power = I^2 R = 4 × 5 = 20 W. / I = 12 / (5 + 1) = 2 A। V_term = 2 × 5 = 10 V। शक्ति = I^2 R = 4 × 5 = 20 W।
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Explain why bulbs in parallel in a domestic circuit do not affect each other’s brightness significantly when one bulb is switched off. / घरेलू सर्किट में बल्ब समानांतर जुड़ने पर जब एक बल्ब बंद किया जाता है तो अन्य बल्बों की चमक पर प्रभावित क्यों नहीं होता, स्पष्ट कीजिए।
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In parallel each bulb has the full supply voltage across it so its current and power remain unchanged when another branch is opened; only total current from supply changes. Hence brightness of other bulbs is unaffected. / समानांतर में प्रत्येक बल्ब पर पूरा आपूर्ति वोल्टेज लगता है, अतः किसी शाखा के खुलने पर उसकी धारा और शक्ति अपरिवर्तित रहती है; केवल कुल आपूर्ति धारा बदलती है। इसलिए अन्य बल्बों की चमक प्रभावित नहीं होती।
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Describe an experimental method using a potentiometer to measure the emf of a cell. / किसी सेल का emf मापने के लिये पोटेंशियोमीटर का उपयोग करते हुए प्रयोगात्मक विधि बताइए।
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Connect a long uniform wire to a stable driving cell establishing a potential gradient. Connect the test cell in series with a galvanometer and jockey; move jockey along wire until galvanometer shows null. Measure balancing length l; using known potential gradient k (found from a standard cell) find emf ε = k l. Ensure no current drawn from test cell at null and maintain constant driving current. Precautions: prevent heating, use sensitive galvanometer, avoid contact resistance. / लंबी समान तार को स्थिर ड्राइविंग सेल से जोड़कर पोटेंशियल ग्रेडियंट बनाइए। परीक्षण सेल को गैल्वानोमीटर और जॉकी के साथ जोड़ीए; जॉकी को तार पर घुमा कर उस बिंदु पर लाइए जहाँ गैल्वानोमीटर शून्य दिखाए। संतुलन लंबाई l मापिए; मानक सेल से ज्ञात ग्रेडियंट k के प्रयोग से emf ε = k l प्राप्त कीजिए। नल अवस्था पर परीक्षण सेल से धारा न निकलती हो, ड्राइविंग धारा स्थिर रखिए। सावधानियाँ: ताप उत्पन्न न होने दें, संवेदनशील गैल्वानोमीटर का प्रयोग करें, संपर्क प्रतिरोध से बचें।
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A 10 Ω resistor dissipates 50 W. Find current through and voltage across it. / एक 10 Ω प्रतिरोध 50 W शक्ति नष्ट कर रहा है। उसमें प्रवाहित धारा और उस पर वोल्टेज ज्ञात कीजिए।
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Use P = V^2 / R so V = √(P R) = √(50 × 10) = √500 ≈ 22.36 V. Current I = P / V = 50 / 22.36 ≈ 2.24 A. Alternatively I = √(P / R) = √(50/10) = √5 ≈ 2.236 A. / P = V^2 / R से V = √(50 × 10) = √500 ≈ 22.36 V। धारा I = P / V = 50 / 22.36 ≈ 2.24 A। वैकल्पिक I = √(P / R) = √(50/10) ≈ 2.236 A।
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Derive the condition for maximum power transfer from a source with emf ε and internal resistance r to a load R. / emf ε और आंतरिक प्रतिरोध r वाले स्रोत से भार R को अधिकतम शक्ति हस्तांतरण की शर्त निकालिए।
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Power to load P = I^2 R where I = ε / (R + r). So P = (ε^2 R) / (R + r)^2. Differentiate w.r.t R and set dP/dR = 0: ε^2[(R + r)^2 - 2R(R + r)]/(R + r)^4 = 0 → (R + r) - 2R = 0 → r - R = 0 so R = r. Thus maximum power when load equals internal resistance. / P = (ε^2 R)/(R + r)^2 पर R के सापेक्ष अवकलन शून्य करने पर शर्त R = r मिलती है, अतः भार R का मान आंतरिक प्रतिरोध के बराबर होने पर शक्ति अधिकतम होती है।
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Two resistors 4 Ω and 12 Ω are connected in parallel across a 24 V battery. Calculate total current drawn from battery and current through each resistor. / 4 Ω और 12 Ω वाले दो प्रतिरोध समानांतर में 24 V बैटरी से जुड़े हैं। बैटरी से ली जाने वाली कुल धारा तथा प्रत्येक प्रतिरोध में प्रवाहित धारा ज्ञात कीजिए।
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Voltage across each = 24 V. Currents: I1 = 24/4 = 6 A; I2 = 24/12 = 2 A. Total I = 8 A. / प्रत्येक पर वोल्टेज 24 V है। I1 = 24/4 = 6 A; I2 = 24/12 = 2 A। कुल I = 8 A।
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Explain briefly why drift velocity of electrons is small though the electric signal travels fast. / इलेक्ट्रिक सिग्नल तीव्रता से पहुँचने के बावजूद इलेक्ट्रॉनों की ड्रिफ्ट वेग छोटी क्यों होती है, संक्षेप में स्पष्ट कीजिए।
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Drift velocity is average slow motion of electrons due to field between collisions; typical v_d is mm s^-1 because electrons frequently collide with lattice ions. The electric signal propagates as an electromagnetic wave at near light speed through the conductor's field, not by individual electron movement; hence signal is fast while electron drift is slow. / ड्रिफ्ट वेग इलेक्ट्रॉनों की टकराहटों के कारण होने वाली औसत धीमी गति है और सामान्यतः मिलीमीटर प्रति सेकंड के क्रम की होती है। जबकि संकेत एक विद्युत-चुंबकीय तरंग के रूप में, जो चालक में बहुत तेज़ी से फैलती है, पहुँचता है; इसलिए सिग्नल तीव्र किन्तु इलेक्ट्रॉन की ड्रिफ्ट धीमी होती है।
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A circuit contains three resistors 2 Ω, 3 Ω and 6 Ω connected in series with a 11 V battery. Find potential drop across each resistor. / एक सर्किट में 2 Ω, 3 Ω और 6 Ω के तीन प्रतिरोध 11 V बैटरी के साथ श्रृंखला में जुड़े हैं। प्रत्येक प्रतिरोध पर विभव गिरावट ज्ञात कीजिए।
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Total R = 2 + 3 + 6 = 11 Ω so I = 11 V / 11 Ω = 1 A. Voltage drops: V1 = I×2 = 2 V; V2 = 3 V; V3 = 6 V. They sum to 11 V. / कुल R = 11 Ω अतः I = 1 A। V1 = 2 V; V2 = 3 V; V3 = 6 V। योग 11 V देता है।
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