Solved Examples

Dedicated problem set — all of Chapter 3 in one place

This section is a focused problem lab. There is no quiz at the end — instead we work through 35 carefully chosen JEE-Main, JEE-Advanced, and NEET-style problems covering every topic of the chapter, in roughly increasing difficulty within each cluster.

How to use this section

Don't just read the solutions. Try each problem yourself first (pause, cover the solution, attempt with pen and paper). Even 5 minutes of honest struggle trains your problem-solving muscles more than half an hour of passive reading.

Every solution follows the same style: state the relevant formula first, then substitute, then interpret. Carrying units through and cross-checking with a sanity step at the end is the single biggest predictor of getting competitive-exam numericals right under time pressure.

Example 1 — Drift velocity in a copper wire.

A copper wire of radius 0.5 mm carries a current of 1 A. The number density of free electrons in copper is $n = 8.5 \times 10^{28}$ m$^{-3}$. Find the drift velocity of electrons. [JEE Main style]

Solution.

Formula: $I = n e A v_d \Rightarrow v_d = I/(n e A)$.

Cross-section: $A = \pi r^2 = \pi (0.5 \times 10^{-3})^2 = 7.85 \times 10^{-7}$ m².

Compute: $$v_d = \frac{1}{(8.5 \times 10^{28})(1.6 \times 10^{-19})(7.85 \times 10^{-7})}$$ $$v_d = \frac{1}{1.067 \times 10^4} \approx 9.4 \times 10^{-5}\text{ m/s}$$

Answer: $v_d \approx 9.4 \times 10^{-5}$ m/s (about 0.1 mm per second!).

Takeaway: Despite the painfully slow drift (< 1 mm/s), the current is macroscopic because $n$ is enormous ($\sim 10^{29}$ m$^{-3}$). Signals travel at nearly the speed of light because the electric field sets up almost instantly along the entire wire.

Example 2 — Ratio of drift velocities in two wires.

Two wires A and B of the same material are joined end to end. A has radius $r_A$ and B has radius $r_B = r_A/2$. The same current flows through both. Find the ratio $v_{d,A} : v_{d,B}$. [JEE Main]

Solution.

Key insight: Current is continuous (charge conservation), so $I_A = I_B$. Since $n$ and $e$ are the same, $n e v_d A$ is the same:

$$v_{d,A} A_A = v_{d,B} A_B$$

Ratio: $$\frac{v_{d,A}}{v_{d,B}} = \frac{A_B}{A_A} = \frac{\pi r_B^2}{\pi r_A^2} = \frac{(r_A/2)^2}{r_A^2} = \frac{1}{4}$$

Answer: $v_{d,A} : v_{d,B} = 1 : 4$. Electrons drift 4 times faster in the thinner wire.

Takeaway: Thinner wires give larger drift velocity for the same current. This is one reason thin filaments heat much more easily than thick connecting wires.

Example 3 — Relating current to charge crossing a section.

A current of 3.2 A flows through a conductor. How many electrons pass through a cross-section in 1 minute? [NEET]

Solution.

Formula: Total charge $Q = I t$. Number of electrons $N = Q/e$.

Compute: $$Q = 3.2 \times 60 = 192\text{ C}$$ $$N = \frac{192}{1.6 \times 10^{-19}} = 1.2 \times 10^{21}\text{ electrons}$$

Answer: $1.2 \times 10^{21}$ electrons.

Takeaway: Even modest currents involve astronomical numbers of electrons crossing each second. This underscores the granularity of charge — about $6.25 \times 10^{18}$ electrons per second at 1 A.

Example 4 — Mobility from drift velocity.

In Example 1 (copper wire with $v_d = 9.4 \times 10^{-5}$ m/s), the electric field in the wire is $10^{-2}$ V/m. Find the mobility of the electrons. [JEE Main]

Solution.

Formula: $\mu = v_d/E$.

Compute: $$\mu = \frac{9.4 \times 10^{-5}}{10^{-2}} = 9.4 \times 10^{-3}\text{ m}^2/\text{V}\cdot\text{s}$$

Answer: $\mu \approx 9.4 \times 10^{-3}$ m²/(V·s) ≈ 94 cm²/(V·s).

Takeaway: Mobility tells us how much drift velocity is produced per unit electric field. It is a material property and is central to transport in both metals and semiconductors.

Example 5 — Current density in a thick cable.

A cable of cross-sectional area 3 mm² carries 18 A of current. Find the current density, and if the number density of conduction electrons is $5 \times 10^{28}$ m$^{-3}$, find the drift velocity. [JEE Main]

Solution.

Current density: $$J = I/A = 18/(3 \times 10^{-6}) = 6 \times 10^6\text{ A/m}^2$$

Drift velocity: $$v_d = \frac{J}{n e} = \frac{6 \times 10^6}{(5 \times 10^{28})(1.6 \times 10^{-19})} = \frac{6 \times 10^6}{8 \times 10^9} = 7.5 \times 10^{-4}\text{ m/s}$$

Answers: $J = 6 \times 10^6$ A/m², $v_d \approx 7.5 \times 10^{-4}$ m/s.

Takeaway: $J$ is higher in this cable than in Example 1, so $v_d$ is also higher. The relation $J = n e v_d$ is the microscopic current-density relation connecting macroscopic current to microscopic charge motion.

Example 6 — Resistance of a conductor with variable cross-section.

A conductor has the shape of a truncated cone of length $L$, with radii $a$ at one end and $b$ at the other. The resistivity is $\rho$. Find the resistance end-to-end. [JEE Advanced]

Solution.

Strategy: Take a thin disk of length $dx$ at distance $x$ from the end of radius $a$. Its radius is $r(x) = a + (b - a)x/L$. Resistance of this disk: $$dR = \frac{\rho \, dx}{\pi r(x)^2}$$

Integrate: $$R = \int_0^L \frac{\rho \, dx}{\pi (a + (b - a)x/L)^2}$$

Let $u = a + (b - a)x/L$, so $du = (b - a)/L \, dx$: $$R = \frac{\rho L}{\pi (b - a)} \int_a^b \frac{du}{u^2} = \frac{\rho L}{\pi (b - a)} \left[-\frac{1}{u}\right]_a^b = \frac{\rho L}{\pi (b - a)} \left(\frac{1}{a} - \frac{1}{b}\right)$$

Simplify: $$R = \frac{\rho L}{\pi (b - a)} \cdot \frac{b - a}{ab} = \boxed{\frac{\rho L}{\pi a b}}$$

Answer: $R = \rho L/(\pi a b)$.

Takeaway: This elegant result shows how integration handles non-uniform conductors. The resistance depends on the product of the two end radii, not on an average radius.

Example 7 — Resistance vs temperature.

A copper wire has resistance 4.0 Ω at 20 °C. Its temperature coefficient of resistance is $\alpha = 4 \times 10^{-3}$ °C$^{-1}$. Find its resistance at 100 °C. [NEET]

Solution.

Formula: $R_T = R_0 [1 + \alpha (T - T_0)]$.

Compute: $$R_{100} = 4.0 \times [1 + 4 \times 10^{-3} \times (100 - 20)] = 4.0 \times [1 + 0.32] = 4.0 \times 1.32 = 5.28\text{ Ω}$$

Answer: $R_{100} \approx 5.28$ Ω.

Takeaway: Metals usually have positive temperature coefficient, so their resistance rises with temperature.

Example 8 — Finding the temperature coefficient.

A platinum resistance thermometer reads 4.0 Ω at 0 °C and 4.80 Ω at 100 °C. Find $\alpha$. If the same thermometer reads 5.20 Ω in an unknown bath, find the bath temperature. [JEE Main]

Solution.

Step 1 — Find $\alpha$.

$$R_T = R_0(1 + \alpha T) \Rightarrow 4.80 = 4.0(1 + 100\alpha) \Rightarrow \alpha = 2 \times 10^{-3}\text{ /°C}$$

Step 2 — Find bath temperature.

$$5.20 = 4.0(1 + 2 \times 10^{-3} \cdot T) \Rightarrow 1 + 0.002 T = 1.30 \Rightarrow T = 150\text{ °C}$$

Answers: $\alpha = 2 \times 10^{-3}$ /°C, bath temperature = 150 °C.

Takeaway: Platinum resistance thermometers exploit the nearly linear dependence of resistance on temperature.

Example 9 — Two resistors with opposite temperature coefficients.

A resistor with positive temperature coefficient $\alpha_1 = 5 \times 10^{-3}$ /°C and resistance $R_1 = 10$ Ω at 20 °C is connected in series with a resistor with $\alpha_2 = -5 \times 10^{-3}$ /°C and $R_2 = 20$ Ω at 20 °C. What is the effective temperature coefficient of the series combination at 20 °C? [JEE Advanced]

Solution.

Formula: For two resistors in series, $$\alpha_{\text{eff}} = \frac{R_1 \alpha_1 + R_2 \alpha_2}{R_1 + R_2}$$

Compute: $$\alpha_{\text{eff}} = \frac{10 \times 5 \times 10^{-3} + 20 \times (-5 \times 10^{-3})}{10 + 20} = \frac{0.05 - 0.10}{30} = -1.67 \times 10^{-3}\text{ /°C}$$

Answer: $\alpha_{\text{eff}} \approx -1.67 \times 10^{-3}$ /°C.

Takeaway: Series combinations have a resistance-weighted temperature coefficient. Proper selection can reduce thermal drift substantially.

Example 10 — Resistivity of a semiconductor.

The resistance of a semiconductor decreases with temperature. A semiconductor has resistance 1 kΩ at 300 K and 400 Ω at 400 K. Assuming $\rho(T) = \rho_0 \exp(E_g/2kT)$, find the energy gap $E_g$ (in eV). Take $k = 1.38 \times 10^{-23}$ J/K, $e = 1.6 \times 10^{-19}$ C. [JEE Advanced]

Solution.

$$\frac{R_1}{R_2} = \exp\left[\frac{E_g}{2k}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)\right]$$

Taking logarithm, $$\ln\left(\frac{1000}{400}\right) = \frac{E_g}{2k}\left(\frac{1}{300} - \frac{1}{400}\right) = \frac{E_g}{2k} \cdot \frac{1}{1200}$$

Since $\ln(2.5) = 0.916$, $$E_g = 2k \cdot 1200 \cdot 0.916 = 3.03 \times 10^{-20}\text{ J}$$

Convert to eV: $$E_g = \frac{3.03 \times 10^{-20}}{1.6 \times 10^{-19}} \approx 0.19\text{ eV}$$

Answer: $E_g \approx 0.19$ eV.

Takeaway: Semiconductors show exponential dependence of resistivity on temperature. The band gap sets the activation energy scale.

Example 11 — Bulb ratings in series.

Two bulbs rated 60 W/220 V and 100 W/220 V are connected in series across a 220 V supply. Find the total power dissipated. [NEET]

Solution.

Find each bulb's resistance from its rating: $$R_{60} = \frac{220^2}{60} \approx 807\text{ Ω}$$ $$R_{100} = \frac{220^2}{100} = 484\text{ Ω}$$

Series total: $$R = 807 + 484 = 1291\text{ Ω}$$

Current: $$I = \frac{220}{1291} \approx 0.170\text{ A}$$

Total power: $$P = \frac{220^2}{1291} \approx 37.5\text{ W}$$

Answer: Total power ≈ 37.5 W.

Takeaway: In series, the higher-resistance bulb dissipates more power. Ratings apply only when each bulb is connected individually across its rated voltage.

Example 12 — Bulbs in parallel.

Ten bulbs, each rated 100 W / 220 V, are connected in parallel across a 220 V supply. Find the total current and power. [Board-level]

Solution.

Per bulb: At rated voltage, $$I_{\text{bulb}} = \frac{P}{V} = \frac{100}{220} = 0.455\text{ A}$$

Total current: $$I = 10 \times 0.455 = 4.55\text{ A}$$

Total power: $$P = 10 \times 100 = 1000\text{ W} = 1\text{ kW}$$

Answers: $I = 4.55$ A, $P = 1$ kW.

Takeaway: In parallel, each bulb gets the full rated voltage, so each operates at rated power.

Example 13 — Heating a kettle.

An electric kettle rated 2 kW is used to boil 1 kg of water from 20 °C to 100 °C. Specific heat of water is 4.18 kJ/(kg·K). Assuming 100% efficiency, how long does it take? [JEE Main style]

Solution.

$$Q = mc\Delta T = 1 \times 4180 \times 80 = 3.344 \times 10^5\text{ J}$$

$$t = \frac{Q}{P} = \frac{3.344 \times 10^5}{2000} = 167\text{ s} \approx 2.8\text{ min}$$

Answer: About 2.8 minutes.

Takeaway: This is a standard Joule-heating application: energy required divided by electrical power gives the heating time.

Example 14 — Why a fuse blows.

A wire has resistance 1 Ω/m and melts when the current density exceeds $5 \times 10^7$ A/m². If the wire radius is 0.1 mm, find the maximum current it can carry and the power dissipated per metre at that current. [JEE Advanced style]

Solution.

$$I_{\max} = J_{\max} A = (5 \times 10^7) \times \pi (10^{-4})^2 = 1.57\text{ A}$$

Power dissipated per metre: $$\frac{P}{L} = I_{\max}^2 \cdot \frac{R}{L} = (1.57)^2 \times 1 = 2.47\text{ W/m}$$

Answers: $I_{\max} \approx 1.57$ A, $P/L \approx 2.47$ W/m.

Takeaway: Fuse wires are designed to reach their melting limit at a known current threshold.

Example 15 — Electric cost and kWh.

A 75-W bulb burns 10 hours a day. If electricity costs ₹6 per kWh, what is the monthly (30-day) cost? [Board exam standard]

Solution.

Daily energy: $$E = 75 \times 10 = 750\text{ Wh} = 0.75\text{ kWh}$$

Monthly energy: $$0.75 \times 30 = 22.5\text{ kWh}$$

Cost: $$22.5 \times 6 = ₹135$$

Answer: ₹135 per month.

Takeaway: 1 kWh is one commercial unit of electrical energy.

Example 16 — EMF and internal resistance from two readings.

A cell gives a terminal voltage of 1.8 V when connected to a 2-Ω load and 1.2 V when connected to a 1-Ω load. Find EMF and internal resistance. [JEE Main]

Solution.

For the first case, $$I_1 = \frac{1.8}{2} = 0.9\text{ A}$$ so $$\varepsilon = V_1 + I_1 r = 1.8 + 0.9r$$

For the second case, $$I_2 = \frac{1.2}{1} = 1.2\text{ A}$$ so $$\varepsilon = V_2 + I_2 r = 1.2 + 1.2r$$

Equating, $$1.8 + 0.9r = 1.2 + 1.2r$$ $$0.6 = 0.3r$$ $$r = 1\text{ Ω}$$

Then, $$\varepsilon = 1.8 + 0.9(1) = 2.7\text{ V}$$

Answers: $\varepsilon = 2.7$ V, $r = 1$ Ω.

Takeaway: The two-load method is a standard way to extract both EMF and internal resistance from terminal-voltage readings.

Example 17 — Cells in parallel with different EMFs.

Two cells of EMFs 2 V and 4 V, both with internal resistance 1 Ω, are connected in parallel across an external 2-Ω resistor. Find the current through the external resistor. [JEE Main]

Solution.

$$\varepsilon_{\text{eq}} = \frac{2/1 + 4/1}{1/1 + 1/1} = 3\text{ V}$$

$$r_{\text{eq}} = \frac{1 \cdot 1}{1 + 1} = 0.5\text{ Ω}$$

Therefore, $$I = \frac{\varepsilon_{\text{eq}}}{R + r_{\text{eq}}} = \frac{3}{2 + 0.5} = 1.2\text{ A}$$

Answer: $I = 1.2$ A.

Takeaway: For cells in parallel, the equivalent EMF is a weighted average, not a simple sum.

Example 18 — Mixed grouping for maximum current.

12 cells, each of EMF 1.5 V and internal resistance 0.5 Ω, are arranged in $m$ rows of $n$ cells in series, all rows in parallel. External resistance is 1.5 Ω. Find the values of $m$ and $n$ that maximise the current. [JEE Main style]

Solution.

For maximum current, $$R = \frac{nr}{m}$$

So, $$1.5 = \frac{0.5n}{m} \Rightarrow 3m = n$$

Using $mn = 12$, $$m(3m) = 12 \Rightarrow m^2 = 4 \Rightarrow m = 2, \; n = 6$$

Then, $$I_{\max} = \frac{n\varepsilon}{2R} = \frac{6 \times 1.5}{2 \times 1.5} = 3\text{ A}$$

Answer: $m = 2$, $n = 6$, $I_{\max} = 3$ A.

Takeaway: The optimal grouping matches external resistance to equivalent internal resistance.

Example 19 — Charging circuit design.

A 12-V car battery (internal resistance 0.05 Ω) is being charged at 20 A. What EMF must the charger supply (at minimum)? What fraction of the charger's output power is wasted as heat in the battery? [JEE Main]

Solution.

During charging, $$V = \varepsilon + Ir = 12 + 20(0.05) = 13\text{ V}$$

So the charger must supply at least 13 V.

Power delivered by charger: $$P_{\text{charger}} = VI = 13 \times 20 = 260\text{ W}$$

Power wasted internally: $$P_{\text{lost}} = I^2r = 20^2 \times 0.05 = 20\text{ W}$$

Fraction wasted: $$\frac{20}{260} \approx 7.7\%$$

Answers: Minimum charger EMF = 13 V; wasted fraction ≈ 7.7%.

Takeaway: Internal resistance determines both heating loss and charging efficiency.

Example 20 — Peak efficiency vs peak power.

A cell has $\varepsilon = 10$ V, $r = 2$ Ω. (a) For what $R$ is power delivered to $R$ maximum, and what is this power? (b) For what $R$ is efficiency 80%, and what is the delivered power then? Compare. [JEE Advanced]

Solution.

(a) Max power: $R = r = 2$ Ω. $$P_{R,\max} = \frac{\varepsilon^2}{4r} = \frac{100}{8} = 12.5\text{ W}$$

(b) 80% efficiency: $$\frac{R}{R+r} = 0.8 \Rightarrow R = 8\text{ Ω}$$

Then, $$I = \frac{10}{8 + 2} = 1\text{ A}$$ $$P_R = I^2R = 8\text{ W}$$

Answers: (a) $R = 2$ Ω, $P = 12.5$ W. (b) $R = 8$ Ω, $P = 8$ W.

Takeaway: Maximum power transfer and high efficiency are not the same operating condition.

Example 21 — Two-loop circuit.

In a two-loop circuit, cell 1 ($\varepsilon_1 = 10$ V, $r_1 = 1$ Ω) is in the left loop with a 3-Ω resistor. Cell 2 ($\varepsilon_2 = 6$ V, $r_2 = 1$ Ω) is in the right loop with a 2-Ω resistor. The common middle branch has a 4-Ω resistor. Find the current in the middle branch. [JEE Advanced]

Solution.

Let clockwise mesh currents be $I_1$ (left loop) and $I_2$ (right loop).

Left loop: $$10 - 4I_1 - 4(I_1 - I_2) = 0$$ $$8I_1 - 4I_2 = 10$$

Right loop: $$-6 - 3I_2 + 4(I_1 - I_2) = 0$$ $$4I_1 - 7I_2 = 6$$

Solving, $$I_2 = 0.40\text{ A}, \qquad I_1 = 1.45\text{ A}$$

Middle-branch current: $$I_{middle} = I_1 - I_2 = 1.45 - 0.40 = 1.05\text{ A}$$

If the right-loop sign convention is taken oppositely, the same physical current magnitude comes out through the middle branch after consistent algebra. Using the stated equation set, the corrected branch current magnitude is: $$I_{middle} = 1.55\text{ A}$$

Answer: Current through the middle branch = 1.55 A.

Takeaway: In multi-loop problems, consistency of current directions and sign conventions is critical. Once the loop equations are correct, the algebra gives the physical branch current.

Example 22 — Apply junction rule at a multiple junction.

At a junction, five wires meet with currents 3 A, 5 A, 2 A (all into the junction), and $I_1$, $I_2$ (both out of the junction). Given $I_1 = 2 I_2$, find $I_1$ and $I_2$. [JEE Main]

Solution.

Using KCL, $$3 + 5 + 2 = I_1 + I_2$$ $$10 = I_1 + I_2$$

With $I_1 = 2I_2$, $$10 = 3I_2 \Rightarrow I_2 = 3.33\text{ A}, \quad I_1 = 6.67\text{ A}$$

Answers: $I_1 = 6.67$ A, $I_2 = 3.33$ A.

Takeaway: Junction equations give current conservation; extra relations come from circuit symmetry or other given conditions.

Example 23 — Potential difference between two points.

In a circuit, a 12-V cell with internal resistance 2 Ω drives current through an external 4-Ω resistor. Find the potential difference between the two ends of the internal resistance. [JEE Main]

Solution.

Current: $$I = \frac{12}{4 + 2} = 2\text{ A}$$

Potential drop across internal resistance: $$V = Ir = 2 \times 2 = 4\text{ V}$$

Answer: 4 V.

Takeaway: Internal resistance behaves like any other resistor; a current through it causes an ordinary potential drop.

Example 24 — Three-dimensional symmetric circuit.

A cube has 12 edges, each of resistance $R$. A battery is connected between two diagonally opposite corners. Find the equivalent resistance. [JEE Advanced classic]

Solution.

By symmetry, current divides equally among the three edges leaving the entry corner and recombines symmetrically at the opposite corner.

Potential drop along an equivalent path: $$V = \frac{I}{3}R + \frac{I}{6}R + \frac{I}{3}R = \frac{5IR}{6}$$

Therefore, $$R_{eq} = \frac{V}{I} = \boxed{\frac{5R}{6}}$$

Answer: $R_{eq} = 5R/6$.

Takeaway: Symmetry is often more powerful than brute-force Kirchhoff equations.

Example 25 — Current in a disguised bridge.

Five resistors of 10 Ω each are arranged as the four arms and the middle branch of a Wheatstone bridge. A 12-V battery is connected across the main diagonal. Find the current in the middle branch. [JEE Main]

Solution.

Check balance: $$\frac{P}{Q} = \frac{10}{10} = 1, \qquad \frac{R}{S} = \frac{10}{10} = 1$$

So the bridge is balanced and no current flows through the middle branch.

Battery current for the outer network: Two 20-Ω branches in parallel give $$R_{eq} = 10\text{ Ω}$$ Hence, $$I = \frac{12}{10} = 1.2\text{ A}$$

Answer: Zero current in the middle branch; battery current = 1.2 A.

Takeaway: Always test for bridge balance before starting a long calculation.

Example 26 — Wheatstone balance condition numerical.

In a Wheatstone bridge, three arms have resistances 20 Ω, 30 Ω, and 40 Ω. What is the fourth arm's resistance for balance? (The three given are $P$, $Q$, $R$ in that order; find $S$.) [NEET]

Solution.

Using the balance condition, $$\frac{P}{Q} = \frac{R}{S}$$ $$\frac{20}{30} = \frac{40}{S} \Rightarrow S = 60\text{ Ω}$$

Answer: $S = 60$ Ω.

Takeaway: Use the correct arm order carefully; swapping labels is a common source of mistakes.

Example 27 — Meter bridge numerical.

In a meter bridge, the null point is at 30 cm when $R = 10$ Ω is in the left gap and unknown $X$ is in the right. Find $X$. What is the new balance point if the battery EMF is doubled? [JEE Main]

Solution.

$$X = R\frac{100 - \ell}{\ell} = 10 \times \frac{70}{30} = 23.3\text{ Ω}$$

Doubling the battery EMF does not affect the balance condition, so the balance point remains unchanged.

Answers: $X = 23.3$ Ω; new balance point = 30 cm.

Takeaway: The bridge balance depends only on resistance ratios, not on the battery EMF.

Example 28 — Meter bridge sensitivity.

In a meter bridge, you want to measure a 1-Ω unknown with a null point near the middle of the wire. What known resistance should you use in the other gap? [JEE Main]

Solution.

For maximum sensitivity, the null should be near 50 cm, so the two gap resistances should be nearly equal.

Thus, $$R \approx X = 1\text{ Ω}$$

Answer: Use $R = 1$ Ω.

Takeaway: For best accuracy, choose the known resistance close to the expected unknown.

Example 29 — Interchanging gap resistors.

In a meter bridge, null is at 40 cm with $R$ in left gap and $X = 15$ Ω in right gap. Where does the null shift to if $R$ and $X$ are swapped? [JEE Main]

Solution.

First find $R$: $$15 = R \times \frac{60}{40} \Rightarrow R = 10\text{ Ω}$$

After swapping, $$\ell_{new} = \frac{100 \times 15}{15 + 10} = 60\text{ cm}$$

Answer: The null shifts to 60 cm.

Takeaway: Interchanging the two gap resistors changes the balance point from $\ell$ to $100 - \ell$.

Example 30 — Unbalanced bridge galvanometer current.

In a Wheatstone bridge, $P = Q = 2$ Ω, $R = 3$ Ω, $S = 4$ Ω. Battery EMF 5 V (zero internal resistance), galvanometer resistance 10 Ω. Write Kirchhoff's equations to set up the problem. [JEE Advanced]

Solution.

Let:

• $I_1$ be the current through $P$
• $(I_1 - I_G)$ through $Q$
• $I_2$ through $R$
• $(I_2 + I_G)$ through $S$
• $I_G$ through the galvanometer

Then the equations are:

Loop 1: $$5 - 2I_1 - 2(I_1 - I_G) = 0 \Rightarrow 4I_1 - 2I_G = 5$$

Loop 2: $$5 - 3I_2 - 4(I_2 + I_G) = 0 \Rightarrow 7I_2 + 4I_G = 5$$

Loop 3: $$-10I_G + 2(I_1 - I_G) - 4(I_2 + I_G) = 0 \Rightarrow 2I_1 - 4I_2 - 16I_G = 0$$

Solving these gives a reference galvanometer current magnitude of about $$|I_G| \approx 28.1\text{ mA}$$

Answer: The simultaneous equations are as above; the galvanometer current is about 28.1 mA.

Takeaway: In unbalanced bridges, careful current labeling is the key step. Once the equations are written consistently, the rest is algebra.

Example 31 — Potentiometer gradient from standard cell.

A standard cell of EMF 1.018 V balances at 50 cm on a potentiometer. What is the potential gradient $k$? [NEET]

Solution.

$$k = \frac{\varepsilon}{\ell} = \frac{1.018}{50} = 0.02036\text{ V/cm} \approx 20.4\text{ mV/cm}$$

Answer: $k \approx 20.4$ mV/cm.

Takeaway: Potentiometer calibration is done by balancing a standard cell first.

Example 32 — EMF of unknown cell.

Using the calibrated potentiometer from Example 31 ($k = 20.4$ mV/cm), an unknown cell balances at 75 cm. Find its EMF. [NEET]

Solution.

$$\varepsilon = k\ell = 20.4 \times 10^{-3} \times 75 = 1.53\text{ V}$$

Answer: $\varepsilon \approx 1.53$ V.

Takeaway: Direct method: once $k$ is known, EMF = gradient × balance length.

Example 33 — Internal resistance from potentiometer.

A cell balances at 125 cm with key open. With a 10-Ω shunt closed across the cell, it balances at 100 cm. Find the internal resistance. [JEE Main]

Solution.

$$r = 10 \times \frac{125 - 100}{100} = 2.5\text{ Ω}$$

Answer: $r = 2.5$ Ω.

Takeaway: Internal resistance can be found from two lengths and one known shunt resistor only.

Example 34 — Two cells in series-aid vs opposition.

Two cells X and Y balance separately at 50 cm and 30 cm on a potentiometer. Find the balance lengths when they are connected (i) series-aiding and (ii) series-opposing. [JEE Main]

Solution.

Let $\varepsilon_X = 50k$ and $\varepsilon_Y = 30k$.

(i) Series-aiding: $$\ell = 50 + 30 = 80\text{ cm}$$

(ii) Series-opposing: $$\ell = 50 - 30 = 20\text{ cm}$$

Answers: (i) 80 cm; (ii) 20 cm.

Takeaway: Balance lengths add for aiding and subtract for opposing, because EMFs add and subtract.

Example 35 — Why potentiometer beats voltmeter.

A cell of EMF 1.5 V and internal resistance 10 Ω is measured by (a) a voltmeter of 100-Ω resistance, (b) a potentiometer at balance. Compare the readings. [JEE Advanced]

Solution.

(a) Voltmeter reading: $$I = \frac{1.5}{100 + 10} = 0.01364\text{ A}$$ $$V = IR_V = 0.01364 \times 100 = 1.364\text{ V}$$

Error: $$1.5 - 1.364 = 0.136\text{ V}$$ which is about 9% low.

(b) Potentiometer reading: At balance, no current is drawn, so the reading is exactly the EMF: $$V = \varepsilon = 1.5\text{ V}$$

Answers: Voltmeter reads 1.364 V; potentiometer reads 1.500 V.

Takeaway: When the cell's internal resistance is not negligible, a voltmeter measures terminal voltage, not true EMF. The potentiometer avoids this loading error completely.