Current Electricity — Easy Practice (Set 1)

Quick Recap — Current, Ohm's Law & Power

• Current $I = \dfrac{Q}{t} = nAev_d$ (unit ampere); the drift velocity is tiny (of order mm/s).
• Ohm's law: $V = IR$; resistance $R = \dfrac{\rho L}{A}$, so $R\propto L$ and $R\propto 1/A$.
• Power: $P = VI = I^2 R = \dfrac{V^2}{R}$ (unit watt); energy $= Pt$ (commercial unit the kWh).
• Meters: an ammeter (low resistance) is connected in series; a voltmeter (high resistance) in parallel.
• Temperature: the resistance of a metal rises with temperature, while a semiconductor's falls.

Beyond-NCERT JEE Formulae

Ohm's law, resistance $R=\rho L/A$ and the series/parallel rules are covered in Sets 1 and 3; this block is the exam-day toolkit for everything beyond them.

1. Kirchhoff's Laws (any non-reducible network)

• Junction (KCL): $\sum I_{in}=\sum I_{out}$ (charge conservation at a node).
• Loop (KVL): $\sum \Delta V=0$ round any closed loop (energy conservation). An EMF is a rise from $-$ to $+$; a resistor gives a drop $IR$ along the current.
• When to use: the moment a network is not pure series/parallel (unbalanced bridge, two or more cells in different loops).
• [JEE Tip] For a two-loop circuit, node-voltage (fix one node at $0$ V and write KCL for the rest) is faster and less error-prone than mesh currents. A branch holding a cell of EMF $E$ pointing toward node $M$ carries $\dfrac{V_M-E}{R}$ out of $M$.

2. Wheatstone Bridge

• Balance condition: $\dfrac{P}{Q}=\dfrac{R}{S}$; the galvanometer then reads zero.
• When to use: null measurement of an unknown resistance, or spotting a balanced bridge hidden in a network.
• [JEE Tip] Balance is independent of the galvanometer resistance, the cell EMF and the cell's position. Always test $\dfrac{P}{Q}=\dfrac{R}{S}$ first: if it holds, delete or short the galvanometer arm and the rest collapses to a plain series-parallel reduction.

3. Meter Bridge (balanced Wheatstone on a 1 m wire)

• Balance: $\dfrac{X}{R}=\dfrac{l}{100-l}$, with $l$ in cm from the left gap.
• Combine with $X=\dfrac{\rho L}{A}$ to convert the balance into the test wire's resistivity.
• [JEE Tip] Aim for balance near the middle (40 to 60 cm), where the fractional error is smallest; interchanging the two gaps and averaging cancels the end-corrections.

4. Potentiometer (draws zero current at balance)

• Potential gradient: $k=\dfrac{V}{L}=\dfrac{I\,R_{AB}}{L}$ (V per m), where the primary current is $I=\dfrac{E_{drv}}{R_s+R_{AB}}$.
• EMF: at balance $E=k\,l$, so it reads the true EMF, not a terminal PD.
• Comparing EMFs: $\dfrac{E_1}{E_2}=\dfrac{l_1}{l_2}$.
• Internal resistance: $r=R\left(\dfrac{l_1-l_2}{l_2}\right)$, where $l_1$ is the open-circuit balance and $l_2$ the balance with a shunt $R$ across the cell.
• [JEE Tip] A potentiometer beats a voltmeter precisely because it draws no current at balance. It works only if the driver EMF exceeds the EMF being measured; a larger series $R_s$ lowers $k$ and lengthens every balance point.

5. Grouping of Cells (identical cells, each EMF $E$, internal $r$; load $R$)

• Series ($n$ in a line): $I=\dfrac{nE}{R+nr}$, best when $R\gg r$.
• Parallel ($m$ cells): $I=\dfrac{E}{R+r/m}$, best when $R\ll r$.
• Mixed ($m$ rows of $n$, total $N=mn$): $I=\dfrac{nE}{R+\dfrac{nr}{m}}$.
• Maximum current when external matches internal resistance: $R=\dfrac{nr}{m}$, giving $I_{max}=\dfrac{mE}{2r}$.
• [JEE Tip] For mismatched cells in parallel use the equivalent-cell (Millman) result $E_{eq}=\dfrac{\sum E_i/r_i}{\sum 1/r_i}$ with $\dfrac{1}{r_{eq}}=\sum\dfrac{1}{r_i}$. A reversed cell subtracts $2E$ from the net EMF, but its $r$ still adds in.

6. Drift Velocity, Mobility, Current Density

• Drift velocity: $v_d=\dfrac{I}{nAe}=\dfrac{eE\tau}{m}$, with $\tau$ the mean relaxation time.
• Mobility: $\mu=\dfrac{v_d}{E}=\dfrac{e\tau}{m}$ (unit m^2 per V per s).
• Current density: $J=\dfrac{I}{A}=nev_d=\sigma E$ (microscopic Ohm's law).
• Conductivity: $\sigma=ne\mu=\dfrac{ne^2\tau}{m}$, and resistivity $\rho=\dfrac{1}{\sigma}$.
• [JEE Tip] $v_d$ is only about mm/s, but $J$ and $\sigma$ are the quantities that carry over between wires of different area. At fixed current $v_d\propto\dfrac{1}{A}$, so a thinner wire means faster drift.

7. RC Circuit Transients

• Time constant: $\tau=RC$.
• Charging: $q=q_0\left(1-e^{-t/RC}\right)$ and $i=\dfrac{V}{R}\,e^{-t/RC}$, with $q_0=CV$.
• Discharging: $q=q_0\,e^{-t/RC}$ and $i=\dfrac{q_0}{RC}\,e^{-t/RC}$.
• Landmarks: one time constant charges to $63\%$ (or decays to $37\%$); half-charge or half-decay takes $t_{1/2}=RC\ln 2\approx0.693\,RC$.
• [JEE Tip] For any resistor network around the capacitor, use Thevenin: $q(t)=CV_{th}\left(1-e^{-t/R_{th}C}\right)$, where $V_{th}$ is the steady-state capacitor voltage and $R_{th}$ is the resistance seen from the capacitor terminals with every ideal EMF shorted. In steady state a capacitor is an open branch.

8. Maximum Power Transfer

• $P_{ext}=\dfrac{E^2 R}{\left(R+r\right)^2}$ is greatest at $R=r$, giving $P_{max}=\dfrac{E^2}{4r}$ at exactly $50\%$ efficiency.
• [JEE Tip] Maximum power is not maximum efficiency. The curve is flat near the peak: at $R=r/2$ or $R=2r$ the load still gets $\dfrac{8}{9}$ of $P_{max}$. If a fixed resistor sits in series with the cell, match $R$ to internal plus fixed, not to $r$ alone.

9. Temperature Coefficient of Resistance

• $R_T=R_0\left(1+\alpha\,\Delta T\right)$, so $\alpha=\dfrac{R_2-R_1}{R_1\left(T_2-T_1\right)}$.
• Metals have $\alpha>0$ (resistance rises with heat); semiconductors and electrolytes have $\alpha<0$.
• [JEE Tip] The reference resistance ($R_0$ or $R_1$) must be the one at the reference temperature: mixing up the baseline is the classic slip. Solve for the rise, then add back any offset such as the starting temperature.

10. Galvanometer Conversion

• To an ammeter (low resistance): a shunt $S=\dfrac{I_g G}{I-I_g}$ in parallel; the finished resistance $\dfrac{GS}{G+S}$ is tiny.
• To a voltmeter (high resistance): a multiplier $R=\dfrac{V}{I_g}-G$ in series; the finished resistance $G+R$ is large.
• The coil always carries the same fraction $\dfrac{I_g}{I}$ of the line current.
• [JEE Tip] Ammeter means a small resistance in parallel; voltmeter means a large resistance in series. Never forget to subtract $G$ in the voltmeter formula, and a real (non-ideal) ammeter's own resistance must be added into the loop.

Solved Examples — Beyond-NCERT Formulae

Example 1 — Wheatstone bridge: balance and equivalent resistance

Given: Bridge arms $P=R_{AB}=10$ ohm, $Q=R_{BC}=20$ ohm and $R=R_{AD}=15$ ohm; find $S=R_{DC}$ for balance and the resistance across $A$ and $C$.

Formula: Balance needs $\dfrac{P}{Q}=\dfrac{R}{S}$. Once balanced the galvanometer arm is dead, so branch $ABC$ (that is $P+Q$) sits in parallel with branch $ADC$ (that is $R+S$).

Working: From balance, $S=\dfrac{Q\,R}{P}=\dfrac{20\times 15}{10}=30$ ohm. The branches are $ABC=30$ ohm and $ADC=15+30=45$ ohm, so $R_{AC}=\dfrac{30\times 45}{30+45}=\dfrac{1350}{75}=18$ ohm.

Answer: $S=30$ ohm and $R_{AC}=18$ ohm.

Example 2 — Meter bridge: unknown resistance and resistivity

Given: A test wire in the left gap balances a standard $R=8$ ohm in the right gap at $l=20$ cm from the left end. The wire is $L=1.0$ m long with cross-section $A=4\times 10^{-7}$ m^2. Find its resistance $X$ and resistivity.

Formula: Balance gives $\dfrac{X}{R}=\dfrac{l}{100-l}$; the geometry then gives $\rho=\dfrac{X\,A}{L}$.

Working: $\dfrac{X}{8}=\dfrac{20}{80}=\dfrac{1}{4}$, so $X=2$ ohm. Then $\rho=\dfrac{2\times 4\times 10^{-7}}{1.0}=8\times 10^{-7}$ ohm m.

Answer: $X=2$ ohm and $\rho=8\times 10^{-7}$ ohm m.

Example 3 — Potentiometer: internal resistance of a cell

Given: On open circuit a cell of EMF $E=1.5$ V balances at $l_1=75$ cm. With a shunt $R=4$ ohm across the cell, the balance falls to $l_2=50$ cm. Find the internal resistance and the loaded terminal PD.

Formula: Balance draws no current, so lengths track potential differences: $r=R\left(\dfrac{l_1-l_2}{l_2}\right)$ and $V=E\,\dfrac{l_2}{l_1}$.

Working: $r=4\times\dfrac{75-50}{50}=4\times\dfrac{25}{50}=2$ ohm. The terminal PD under load is $V=1.5\times\dfrac{50}{75}=1.0$ V.

Answer: $r=2$ ohm and terminal PD $=1.0$ V.

Example 4 — Grouping of cells for maximum current

Given: $N=24$ identical cells, each of EMF $E=1.5$ V and internal resistance $r=0.5$ ohm, are arranged as $m$ parallel rows of $n$ in series (so $mn=24$) to drive a load $R=3$ ohm. Find the arrangement giving the largest current and that current.

Formula: With $m$ rows of $n$, the battery has EMF $nE$ and internal resistance $\dfrac{nr}{m}$, so $I=\dfrac{nE}{R+nr/m}$, which is greatest when $R=\dfrac{nr}{m}$.

Working: Set $R=\dfrac{nr}{m}$: $3=\dfrac{0.5\,n}{m}$ gives $\dfrac{n}{m}=6$, so $n=6m$. Combined with $mn=24$ this gives $6m^2=24$, so $m=2$ rows and $n=12$ in series. Then the battery EMF is $nE=18$ V and its internal resistance $\dfrac{nr}{m}=3$ ohm, so $I=\dfrac{18}{3+3}=3$ A. (Check: $I_{max}=\dfrac{mE}{2r}=\dfrac{2\times 1.5}{2\times 0.5}=3$ A.)

Answer: Two rows of twelve cells, giving $I=3$ A.

Example 5 — RC charging via a Thevenin reduction

Given: A 12 V ideal battery is in series with a $4$ ohm resistor, which feeds a $12$ ohm resistor in parallel with a capacitor $C=4$ microfarad (initially uncharged). Find the final charge and the time to reach half of it.

Formula: Reduce the network around the capacitor to Thevenin form: $q=CV_{th}\left(1-e^{-t/R_{th}C}\right)$, where $V_{th}$ is the steady-state capacitor voltage and $R_{th}$ is the resistance at its terminals with the battery shorted.

Working: In steady state no current enters $C$, so the divider gives $V_{th}=12\times\dfrac{12}{4+12}=9$ V, hence $q_\infty=CV_{th}=4\times 9=36$ microcoulomb. Shorting the battery puts $4$ ohm parallel with $12$ ohm, so $R_{th}=\dfrac{4\times 12}{16}=3$ ohm and $\tau=R_{th}C=3\times 4=12$ microseconds. Half charge occurs at $t=\tau\ln 2=12\times 0.693\approx 8.3$ microseconds.

Answer: $q_\infty=36$ microcoulomb and $t_{1/2}\approx 8.3$ microseconds.

Example 6 — Galvanometer converted to an ammeter (shunt)

Given: A galvanometer of $G=49$ ohm gives full-scale deflection at $I_g=0.02$ A (that is 20 mA). Convert it to an ammeter of range $I=1.0$ A.

Formula: The shunt satisfies $S=\dfrac{I_g G}{I-I_g}$ in parallel; the finished ammeter resistance is $\dfrac{GS}{G+S}$, and the coil carries the fraction $\dfrac{I_g}{I}$ of the line current.

Working: $S=\dfrac{0.02\times 49}{1.0-0.02}=\dfrac{0.98}{0.98}=1.0$ ohm. The ammeter resistance is $\dfrac{49\times 1.0}{49+1.0}=0.98$ ohm, and the coil carries $\dfrac{0.02}{1.0}=2\%$ of the line current.

Answer: Shunt $S=1.0$ ohm in parallel; ammeter resistance $0.98$ ohm; coil takes $2\%$.

Example 7 — Drift velocity, current density and mobility

Given: A copper wire has $A=2\times 10^{-6}$ m^2 and carries $I=3.2$ A, with $n=8\times 10^{28}$ per m^3, $e=1.6\times 10^{-19}$ C and field $E=0.032$ V/m along it.

Formula: Current density $J=\dfrac{I}{A}$; drift speed $v_d=\dfrac{I}{nAe}$; conductivity $\sigma=\dfrac{J}{E}$; mobility $\mu=\dfrac{v_d}{E}$.

Working: $J=\dfrac{3.2}{2\times 10^{-6}}=1.6\times 10^{6}$ A/m^2. The drift speed is $v_d=\dfrac{3.2}{8\times 10^{28}\times 2\times 10^{-6}\times 1.6\times 10^{-19}}=1.25\times 10^{-4}$ m/s. Then $\sigma=\dfrac{1.6\times 10^{6}}{0.032}=5\times 10^{7}$ S/m and $\mu=\dfrac{1.25\times 10^{-4}}{0.032}\approx 3.9\times 10^{-3}$, in m^2 per V per s.

Answer: $J=1.6\times 10^{6}$ A/m^2, $v_d=1.25\times 10^{-4}$ m/s, $\sigma=5\times 10^{7}$ S/m and $\mu\approx 3.9\times 10^{-3}$ m^2 per V per s.

Example 8 — Maximum power transfer with a fixed series resistor

Given: A cell of EMF $E=20$ V and internal resistance $r=2$ ohm has a fixed $R_f=3$ ohm in series, feeding a variable load $R$. Find the load for maximum power in $R$, that power, and the efficiency.

Formula: As far as $R$ can tell, the source has internal resistance $r+R_f$, so the power in $R$ peaks at $R=r+R_f$; then $P=I^2R$ while the total generated power is $EI$.

Working: $R=r+R_f=5$ ohm, so $I=\dfrac{20}{5+5}=2$ A. Then $P_{max}=I^2R=2^2\times 5=20$ W. The total generated is $EI=20\times 2=40$ W, so the efficiency is $\dfrac{20}{40}=50\%$.

Answer: $R=5$ ohm, $P_{max}=20$ W, at $50\%$ efficiency.