Physics 2

Solved exercises on current, DC circuits and the RC circuit

Thirteen exercises with solutions: resistivity, drift velocity, series and parallel, maximum power transfer, Drude model, Kirchhoff's laws, Wheatstone bridge, RC charge and discharge, light bulb, coaxial cable, Hall effect, Thévenin.

ielettronica 25 October 2026 10 min read 0 comments

Exercises on electric current, resistance, DC circuits and the RC circuit of the kind set in Physics 2 courses, in three levels: basic, for resistivity, drift velocity and resistor combinations; intermediate, with the microscopic model, Kirchhoff’s laws, the Wheatstone bridge and RC transients; exam, with complete problems as in written exams. The result is given straight away; the full solution opens with a click. The proofs are in the lesson and the in-depth study linked at the top.

Constants: $e = 1.602\cdot 10^{-19}$ C, $m_e = 9.109\cdot 10^{-31}$ kg; copper: $\rho = 1.7\cdot 10^{-8}$ Ω·m, $n = 8.5\cdot 10^{28}$ m $^{-3}$ .

Basic level

Exercise 1. The resistance of a wire

Compute the resistance of a copper wire 50 m long with a diameter of 1.0 mm. What diameter should an aluminium wire ( $\rho = 2.8\cdot 10^{-8}$ Ω·m) of the same length have to give the same resistance?

Result: 1.08 Ω; 1.28 mm.

Solution

Cross-section: $A = \pi r^2 = \pi(0.5\cdot 10^{-3})^2 = 7.85\cdot 10^{-7}$ m². Resistance: $R = \rho L/A = 1.7\cdot 10^{-8}\cdot 50/7.85\cdot 10^{-7} = 1.08$ Ω.

For the same $R$ with the same length we need $A_{Al}/A_{Cu} = \rho_{Al}/\rho_{Cu}$ , and the diameter goes as the square root: $d_{Al} = 1.0\cdot\sqrt{2.8/1.7} = 1.28$ mm. Despite the larger cross-section, the aluminium wire weighs about half as much (density 2.7 against 8.9 g/cm³): this is why overhead lines are made of aluminium.

Exercise 2. How slow the drift is

A current of 10 A flows in a 2.5 mm² copper cable. Find the drift velocity of the electrons and the time they take to travel 1 m.

Result: $2.9\cdot 10^{-4}$ m/s; about 57 minutes.

Solution

Current density: $J = I/A = 10/2.5\cdot 10^{-6} = 4.0\cdot 10^6$ A/m². Drift velocity: $v_d = \dfrac{J}{ne} = \dfrac{4.0\cdot 10^6}{8.5\cdot 10^{28}\cdot 1.602\cdot 10^{-19}} = 2.9\cdot 10^{-4}$ m/s.

Time for 1 m: $1/2.9\cdot 10^{-4} \approx 3400$ s, almost an hour. The bulb lights up immediately because the field propagates, not the electrons.

Exercise 3. Series and parallel

In the circuit of the figure $E = 12$ V, $R_1 = 6$ Ω, $R_2 = 3$ Ω, $R_3 = 6$ Ω. Find the equivalent resistance and the currents $I$ , $I_2$ , $I_3$ .

Circuit with a source, a series resistor and two parallel resistors — The source E feeds R1 in series with the parallel combination of R2 and R3. The current I flows through R1 and splits into I2 and I3 in the two parallel branches.

Result: $R_{eq} = 8$ Ω; $I = 1.5$ A, $I_2 = 1.0$ A, $I_3 = 0.5$ A.

Solution

Parallel: $R_{23} = \dfrac{R_2R_3}{R_2 + R_3} = \dfrac{18}{9} = 2$ Ω. In series with $R_1$ : $R_{eq} = 8$ Ω.

Total current $I = 12/8 = 1.5$ A. Voltage across the parallel pair: $V_{23} = R_{23}I = 3$ V. So $I_2 = 3/3 = 1.0$ A and $I_3 = 3/6 = 0.5$ A. Check: $I_2 + I_3 = I$ .

Current divider rule: in a parallel combination the current splits in inverse proportion to the resistances.

Exercise 4. Real source and maximum power transfer

A battery has an electromotive force of 12 V and internal resistance 0.5 Ω, and is connected to a 5.5 Ω load. Find the current, the terminal voltage and the power in the load. For which load resistance would the power be maximum, and what would the efficiency be?

Result: 2 A; 11 V; 22 W; maximum for $R = 0.5$ Ω, 72 W, efficiency 50 %.

Solution

$I = \dfrac{E}{R + r} = \dfrac{12}{6} = 2$ A. Terminal voltage $V = E - rI = 11$ V. Power in the load $P = RI^2 = 22$ W.

Power as a function of $R$ : $P(R) = \dfrac{E^2R}{(R + r)^2}$ . Differentiating, $P'(R) = E^2\dfrac{(R + r) - 2R}{(R + r)^3} = E^2\dfrac{r - R}{(R + r)^3}$ , which vanishes for $R = r$ . Then $P_{max} = \dfrac{E^2}{4r} = \dfrac{144}{2} = 72$ W.

Efficiency $\eta = \dfrac{RI^2}{EI} = \dfrac{R}{R + r}$ : with $R = r$ half of the power is lost in the source. This is why maximum transfer is sought for signals (antennas, amplifiers), never in power distribution, where one wants $R \gg r$ .

Load power and efficiency as functions of the ratio between load resistance and internal resistance — The power, in blue, peaks when the load equals the internal resistance; the efficiency, in orange, is one half at that point and rises towards 1 for large loads.

Intermediate level

Exercise 5. The Drude model for silver

Silver has resistivity $1.59\cdot 10^{-8}$ Ω·m and $5.86\cdot 10^{28}$ free electrons per m³. Compute the mean time between collisions and the drift velocity in a 1 mm² wire carrying 1 A.

Result: $\tau = 3.8\cdot 10^{-14}$ s; $v_d = 1.07\cdot 10^{-4}$ m/s.

Solution

From the Drude model (in-depth study) $\sigma = ne^2\tau/m$ , so

$\tau = \frac{m}{ne^2\rho} = \frac{9.109\cdot 10^{-31}}{5.86\cdot 10^{28}\cdot(1.602\cdot 10^{-19})^2\cdot 1.59\cdot 10^{-8}} = 3.8\cdot 10^{-14}\ \text{s}$

Drift velocity: $v_d = \dfrac{J}{ne} = \dfrac{10^6}{5.86\cdot 10^{28}\cdot 1.602\cdot 10^{-19}} = 1.07\cdot 10^{-4}$ m/s, a little more than copper because silver has fewer free electrons per unit volume.

Silver conducts better than copper despite having fewer carriers: the time between collisions is longer ( $3.8$ against $2.5\cdot 10^{-14}$ s).

Exercise 6. Kirchhoff’s laws with two sources

In the circuit of the figure $E_1 = 10$ V, $E_2 = 5$ V, $R_1 = 2$ Ω, $R_2 = 4$ Ω, $R_3 = 4$ Ω. Find the three currents. Does the source $E_2$ deliver or absorb power?

Circuit with two sources and three branches — The left branch contains E1 and R1, the right one E2 and R2, the middle one R3. The three branches are connected between nodes A and B. The arrows show the directions chosen for the currents.

Result: $I_1 = 1.875$ A, $I_2 = -0.3125$ A, $I_3 = 1.5625$ A; $E_2$ absorbs 1.56 W (it is being charged).

Solution

Use nodal analysis: taking B as reference (0 V), the only unknown is the potential $V_A$ . With the directions of the figure:

$I_1 = \dfrac{E_1 - V_A}{R_1}$ , $I_2 = \dfrac{E_2 - V_A}{R_2}$ , $I_3 = \dfrac{V_A}{R_3}$ .

Current law at A: $I_1 + I_2 = I_3$ :

$\frac{10 - V_A}{2} + \frac{5 - V_A}{4} = \frac{V_A}{4}$

Multiplying by 4: $20 - 2V_A + 5 - V_A = V_A$ , so $V_A = 6.25$ V.

Currents: $I_1 = 3.75/2 = 1.875$ A; $I_2 = -1.25/4 = -0.3125$ A; $I_3 = 6.25/4 = 1.5625$ A. Check: $1.875 - 0.3125 = 1.5625$ .

$I_2$ negative means that the current in the right branch flows opposite to the arrow: it enters the positive terminal of $E_2$ . The source absorbs the power $E_2|I_2| = 5\cdot 0.3125 = 1.56$ W: it is a battery being charged.

Balance: $E_1$ delivers $10\cdot 1.875 = 18.75$ W; the resistors dissipate $2\cdot 1.875^2 + 4\cdot 0.3125^2 + 4\cdot 1.5625^2 = 7.03 + 0.39 + 9.77 = 17.19$ W; plus 1.56 W absorbed by $E_2$ makes 18.75 W.

Exercise 7. The Wheatstone bridge

In the bridge of the figure $R_1 = 100$ Ω, $R_2 = 200$ Ω, $R_3 = 150$ Ω. The galvanometer G reads zero. What is $R_x$ ? Prove the balance condition.

Result: $R_x = 300$ Ω.

Solution

If no current flows in the galvanometer, C and D are at the same potential and the two sides of the bridge are two independent dividers, fed by the same voltage $E$ .

Left side: $V_C = E\dfrac{R_3}{R_1 + R_3}$ (relative to the bottom vertex). Right side: $V_D = E\dfrac{R_x}{R_2 + R_x}$ .

$V_C = V_D$ gives $\dfrac{R_3}{R_1 + R_3} = \dfrac{R_x}{R_2 + R_x}$ , i.e. $R_3R_2 + R_3R_x = R_xR_1 + R_xR_3$ , and therefore

$R_x = \frac{R_2R_3}{R_1} = \frac{200\cdot 150}{100} = 300\ \Omega$

The condition depends neither on $E$ nor on the galvanometer resistance: this is why the bridge measures resistances very accurately, and in its unbalanced version it is the circuit of strain gauges and bridge sensors.

Exercise 8. Charging a capacitor

A 47 µF capacitor, initially uncharged, is charged through a 10 kΩ resistor from a 9 V battery. Find the time constant, the initial current, the time to reach 6 V and the energy stored at the end.

Result: $\tau = 0.47$ s; 0.90 mA; 0.52 s; 1.9 mJ.

Solution

$\tau = RC = 10^4\cdot 47\cdot 10^{-6} = 0.47$ s. At the initial instant the capacitor is a short circuit: $i(0) = E/R = 0.90$ mA.

From $v(t) = E\left(1 - e^{-t/\tau}\right)$ : $e^{-t/\tau} = 1 - \frac{6}{9} = \frac{1}{3}$ , so $t = \tau\ln 3 = 0.47\cdot 1.099 = 0.52$ s.

Final energy: $\frac{1}{2}CE^2 = \frac{1}{2}\cdot 47\cdot 10^{-6}\cdot 81 = 1.9$ mJ. The battery supplied twice as much, $CE^2 = 3.8$ mJ: the other half was dissipated in the resistor (lesson).

RC circuit while charging — On the left the circuit; on the right the capacitor voltage rising towards E and the current falling towards zero, with the time constant tau.

Exercise 9. Discharging a capacitor

A 100 µF capacitor charged to 50 V discharges through a 2 kΩ resistor. After how long does the voltage fall to 5 V? How much energy is dissipated in total?

Result: 0.46 s; 0.125 J.

Solution

During discharge $v(t) = V_0e^{-t/\tau}$ , with $\tau = RC = 2000\cdot 10^{-4} = 0.2$ s.

$5 = 50\,e^{-t/\tau}$ gives $t = \tau\ln 10 = 0.2\cdot 2.303 = 0.46$ s. Every time constant the voltage drops to $1/e \approx 37$ %.

Energy: all the stored energy, $\frac{1}{2}CV_0^2 = \frac{1}{2}\cdot 10^{-4}\cdot 2500 = 0.125$ J, ends up as heat in the resistor. Check by integration: $\int_0^\infty\frac{v^2}{R}dt = \frac{V_0^2}{R}\cdot\frac{\tau}{2} = \frac{2500}{2000}\cdot 0.1 = 0.125$ J.

Capacitor discharge — The voltage starts at 50 V and decays exponentially; after one time constant, orange dot, it is about 18 V; at time t1, red dot, it reaches 5 V.

Exam level

Exercise 10. The incandescent bulb

A tungsten-filament bulb is rated 60 W at 230 V. At room temperature (20 °C) the filament measures 70 Ω. With the temperature coefficient $\alpha = 4.5\cdot 10^{-3}$ K $^{-1}$ , estimate the operating temperature, and compare the current at switch-on with the steady-state current.

Result: about 2600 °C; at switch-on 3.3 A against 0.26 A in steady state (about 13 times).

Solution

Hot resistance: $R = V^2/P = 230^2/60 = 882$ Ω.

With the linear law $R = R_{20}(1 + \alpha\,\Delta T)$ : $1 + \alpha\,\Delta T = 882/70 = 12.6$ , so $\Delta T = 11.6/4.5\cdot 10^{-3} \approx 2580$ K and $T \approx 2600$ °C. (The linear law is only an approximation over such a wide range, but the order of magnitude is right: tungsten melts at 3422 °C.)

Currents: in steady state $230/882 = 0.26$ A; at switch-on, with the filament cold, $230/70 = 3.3$ A (rms values). The peak lasts a few milliseconds, until the filament heats up, and it is why bulbs almost always burn out at the moment they are switched on.

Exercise 11. The insulation of a coaxial cable

A coaxial cable 100 m long has an inner conductor of radius 0.50 mm and a braid of radius 1.75 mm, separated by polyethylene ( $\rho = 10^{13}$ Ω·m, $\varepsilon_r = 2.3$ ). Compute the insulation resistance, the leakage current at 1 kV, and the product $RC$ of insulation resistance and cable capacitance.

Result: $2.0\cdot 10^{10}$ Ω; 50 nA; $RC = \rho\varepsilon = 204$ s, independent of geometry.

Solution

Radial resistance (in-depth study): $R = \dfrac{\rho}{2\pi L}\ln\dfrac{b}{a} = \dfrac{10^{13}}{2\pi\cdot 100}\ln 3.5 = 2.0\cdot 10^{10}$ Ω. Current at 1 kV: $10^3/2.0\cdot 10^{10} = 50$ nA.

Capacitance of the coaxial cable: $C = \dfrac{2\pi\varepsilon L}{\ln(b/a)} = \dfrac{2\pi\cdot 2.3\cdot 8.854\cdot 10^{-12}\cdot 100}{1.253} = 10.2$ nF.

Product: $RC = \dfrac{\rho}{2\pi L}\ln\dfrac{b}{a}\cdot\dfrac{2\pi\varepsilon L}{\ln(b/a)} = \rho\varepsilon = 10^{13}\cdot 2.3\cdot 8.854\cdot 10^{-12} = 204$ s.

The result $RC = \rho\varepsilon$ holds for any pair of electrodes immersed in a homogeneous medium, because resistance and capacitance depend on geometry in inverse ways. It is the time in which a capacitor with that dielectric discharges by itself through leakage.

Radial leakage current in a coaxial cable — The current flows from the inner conductor to the braid through the insulation, in concentric cylindrical shells.

Exercise 12. Measuring carrier density with the Hall effect

A semiconductor strip 0.20 mm thick carries 5.0 mA in a perpendicular magnetic field of 0.50 T. A Hall voltage of 25 mV is measured, with the sign indicating negative carriers. Find the carrier density and compare it with copper.

Result: $n = 3.1\cdot 10^{21}$ m $^{-3}$ , about 27 million times less than copper.

Solution

From the Hall effect formula (in-depth study) $V_H = \dfrac{IB}{nqt}$ :

$n = \frac{IB}{e\,t\,V_H} = \frac{5.0\cdot 10^{-3}\cdot 0.50}{1.602\cdot 10^{-19}\cdot 2.0\cdot 10^{-4}\cdot 0.025} = 3.1\cdot 10^{21}\ \text{m}^{-3}$

Ratio to copper: $8.5\cdot 10^{28}/3.1\cdot 10^{21} \approx 2.7\cdot 10^7$ . It is precisely because $n$ is small that the Hall voltage is large and measurable: in copper, with the same data, it would be about 1 nV.

The sign says that the carriers are electrons: it is an n-type semiconductor.

Hall effect in a strip — The Lorentz force pushes the carriers onto one edge of the strip, and the Hall voltage forms between the two edges.

Exercise 13. An RC circuit with Thévenin’s theorem

In the circuit of the figure $E = 12$ V, $R_1 = 4$ kΩ, $R_2 = 12$ kΩ, $C = 10$ µF, and the capacitor is initially uncharged. Find the capacitor voltage as a function of time and the current delivered by the source at the beginning and at the end.

RC circuit with two resistors and its Thévenin equivalent — On the left the source E feeds R1 in series, then R2 and the capacitor C in parallel. On the right the same circuit seen from the capacitor: a source V th with R th in series.

Result: $v_C(t) = 9\left(1 - e^{-t/30\,\text{ms}}\right)$ V; source current 3.0 mA at the start and 0.75 mA in steady state.

Solution

Seen from the capacitor terminals, the rest of the circuit is a divider. Thévenin: open-circuit voltage $V_{th} = E\dfrac{R_2}{R_1 + R_2} = 12\cdot\dfrac{12}{16} = 9$ V; resistance with the source turned off (short circuit) $R_{th} = R_1 \parallel R_2 = \dfrac{4\cdot 12}{16} = 3$ kΩ.

The equivalent circuit is a simple RC: $\tau = R_{th}C = 3\cdot 10^3\cdot 10^{-5} = 30$ ms, and $v_C(t) = 9\left(1 - e^{-t/\tau}\right)$ V. The capacitor does not reach 12 V but 9 V, the steady-state voltage across $R_2$ .

Source current: at the start the capacitor is a short circuit that bypasses $R_2$ , so $i = E/R_1 = 3.0$ mA. In steady state the capacitor is an open circuit, and $i = E/(R_1 + R_2) = 0.75$ mA.

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Basic level

Exercise 1. The resistance of a wire

Exercise 2. How slow the drift is

Exercise 3. Series and parallel

Exercise 4. Real source and maximum power transfer

Intermediate level

Exercise 5. The Drude model for silver

Exercise 6. Kirchhoff’s laws with two sources

Exercise 7. The Wheatstone bridge

Exercise 8. Charging a capacitor

Exercise 9. Discharging a capacitor

Exam level

Exercise 10. The incandescent bulb

Exercise 11. The insulation of a coaxial cable

Exercise 12. Measuring carrier density with the Hall effect

Exercise 13. An RC circuit with Thévenin’s theorem

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