1.P — Worked Problems: Circuit Theory

Sixteen problems covering everything in Part 1. Every solution shows the arithmetic, and most end with a check — because in circuit work the power balance or a KCL sum will catch a mistake that a plausible-looking answer will not. Attempt each one before opening the solution.

Problem 1 — Charge, current and battery capacity

A phone battery is rated 4500 mAh at 3.85 V. (a) How many coulombs does it hold? (b) How much energy in joules and watt-hours? (c) If the phone draws an average of 220 mA, how long does it last?

Solution

(a) Convert to amp-seconds. 4500 mAh = 4.5 Ah, and one hour is 3600 s:

Q = 4.5 \times 3600 = 16{,}200\ \text{C}

(b) Energy is charge times voltage, since voltage is joules per coulomb:

E = QV = 16{,}200 \times 3.85 = 62{,}370\ \text{J}

In watt-hours, 4.5\ \text{Ah}\times3.85\ \text{V} = 17.3 Wh. Check: 17.3 \times 3600 = 62{,}300 J ✓ (small rounding).

(c)

t = \frac{4500\ \text{mAh}}{220\ \text{mA}} = 20.5\ \text{hours}

What to notice. The 3.85 V is an average; a lithium cell actually runs from about 4.2 V down to 3.0 V as it empties, so the amp-hour figure is the honest one and the watt-hour figure is an approximation. This is why battery banks are advertised in mAh but the useful comparison is Wh.

Problem 2 — Resistivity and cable sizing

A copper cable of 2.5 mm² cross-section runs 30 m to a load drawing 16 A. Copper's resistivity is $1.68\times10^{-8}\ \Omega!\cdot!$m. Find the cable's resistance, the voltage lost in it, and the power wasted. Then repeat for 1.5 mm² cable.

Solution

Resistance of the 2.5 mm² cable. The current goes out and comes back, so the conductor length is 2\times30 = 60 m. Area is 2.5\ \text{mm}^2 = 2.5\times10^{-6}\ \text{m}^2:

R = \rho\frac{\ell}{A} = 1.68\times10^{-8}\times\frac{60}{2.5\times10^{-6}} = 1.68\times10^{-8}\times2.4\times10^{7} = 0.403\ \Omega

Voltage lost:

V = IR = 16 \times 0.403 = 6.45\ \text{V}

Power wasted:

P = I^2R = 256 \times 0.403 = 103\ \text{W}

For 1.5 mm², the area is 0.6 times as large, so the resistance is 1/0.6 = 1.667 times bigger:

R = 0.672\ \Omega, \quad V = 10.7\ \text{V}, \quad P = 172\ \text{W}

What to notice. On a 230 V supply, 6.45 V lost is 2.8%, which is within the usual 3% wiring standard. The thinner cable loses 4.7%, outside it, and dissipates 172 W inside a wall — enough to be a genuine fire risk in insulation. This is the entire reason cable size is regulated by current rating, and the I^2R dependence is why the penalty grows so fast.

Problem 3 — Reduce and solve a ladder

A 24 V source drives R_1 = 6\ \Omega in series with the parallel pair R_2 = 12\ \Omega and R_3 = 4\ \Omega, which is in series with the parallel pair R_4 = 10\ \Omega and R_5 = 15\ \Omega. Find the total current, then the current in every resistor.

Solution

Step 1 — collapse the parallel pairs.

R_{23} = \frac{12\times4}{16} = 3\ \Omega, \qquad R_{45} = \frac{10\times15}{25} = 6\ \Omega

Both smaller than their smallest member ✓.

Step 2 — total.

R_{total} = 6+3+6 = 15\ \Omega, \qquad I_{total} = \frac{24}{15} = 1.6\ \text{A}

Step 3 — expand back out.

Voltage across the first pair: 1.6\times3 = 4.8 V. So

I_2 = \frac{4.8}{12} = 0.4\ \text{A}, \qquad I_3 = \frac{4.8}{4} = 1.2\ \text{A}

Voltage across the second pair: 1.6\times6 = 9.6 V. So

I_4 = \frac{9.6}{10} = 0.96\ \text{A}, \qquad I_5 = \frac{9.6}{15} = 0.64\ \text{A}

Checks. 0.4+1.2 = 1.6 ✓, 0.96+0.64 = 1.6 ✓. Voltage sum: 1.6\times6 + 4.8 + 9.6 = 9.6+4.8+9.6 = 24 ✓.

Power check. Source: 24\times1.6 = 38.4 W. Absorbed: 1.6^2\times6 = 15.36, plus 4.8^2/3 = 7.68, plus 9.6^2/6 = 15.36. Total 38.4 W ✓.

Problem 4 — Nodal analysis with two sources

A 15 V source feeds node A through 5\ \Omega. Node A connects to node B through 10\ \Omega. Node B has 20\ \Omega to ground and a 2 A current source injecting into it. Node A also has 8\ \Omega to ground. Find V_A and V_B.

Solution

Node A, all currents leaving:

\frac{V_A-15}{5} + \frac{V_A}{8} + \frac{V_A-V_B}{10} = 0

Multiply by 40:

8(V_A-15) + 5V_A + 4(V_A-V_B) = 0

8V_A - 120 + 5V_A + 4V_A - 4V_B = 0

17V_A - 4V_B = 120 \qquad (1)

Node B, all currents leaving, with the 2 A source counted as an injection on the right:

\frac{V_B-V_A}{10} + \frac{V_B}{20} = 2

Multiply by 20:

2(V_B-V_A) + V_B = 40

-2V_A + 3V_B = 40 \qquad (2)

From (2), V_A = (3V_B-40)/2. Substitute into (1):

17\cdot\frac{3V_B-40}{2} - 4V_B = 120

Multiply by 2:

51V_B - 680 - 8V_B = 240

43V_B = 920 \quad\Rightarrow\quad V_B = 21.4\ \text{V}

Then V_A = (64.2-40)/2 = 12.1\ \text{V}.

Check node B: (21.4-12.1)/10 + 21.4/20 = 0.93+1.07 = 2.0 ✓.

What to notice. V_B is higher than the 15 V supply, because the current source is pushing current into the network and current is flowing backwards through the 10 Ω resistor towards node A. Nothing is wrong; that is what an injected current does.

Problem 5 — Mesh analysis with three meshes

Three meshes in a row. Mesh 1 has a 30 V source and R_a = 6\ \Omega; mesh 1 and 2 share R_b = 12\ \Omega; mesh 2 has R_c = 8\ \Omega of its own; mesh 2 and 3 share R_d = 4\ \Omega; mesh 3 has R_e = 10\ \Omega and a 20 V source opposing. Find the three mesh currents.

Solution

Using the by-inspection pattern with all currents clockwise:

Mesh 1: (6+12)I_1 - 12I_2 = 30, so 18I_1 - 12I_2 = 30, or dividing by 6: 3I_1 - 2I_2 = 5 (1)

Mesh 2: -12I_1 + (12+8+4)I_2 - 4I_3 = 0, so -12I_1 + 24I_2 - 4I_3 = 0, or dividing by 4: -3I_1 + 6I_2 - I_3 = 0 (2)

Mesh 3: -4I_2 + (4+10)I_3 = -20, so -4I_2 + 14I_3 = -20, or dividing by 2: -2I_2 + 7I_3 = -10 (3)

From (3): I_3 = (2I_2-10)/7. Substitute into (2):

-3I_1 + 6I_2 - \frac{2I_2-10}{7} = 0

Multiply by 7:

-21I_1 + 42I_2 - 2I_2 + 10 = 0 \;\Rightarrow\; -21I_1 + 40I_2 = -10 \qquad (4)

From (1): I_1 = (5+2I_2)/3. Substitute into (4):

-21\cdot\frac{5+2I_2}{3} + 40I_2 = -10

-7(5+2I_2) + 40I_2 = -10

-35 - 14I_2 + 40I_2 = -10

26I_2 = 25 \;\Rightarrow\; I_2 = 0.962\ \text{A}

Then I_1 = (5+1.923)/3 = 2.308 A, and I_3 = (1.923-10)/7 = -1.154 A.

What the negative means. I_3 flows anticlockwise, which is correct — the 20 V source in mesh 3 is pushing that way.

Check mesh 1: 18(2.308) - 12(0.962) = 41.5 - 11.5 = 30 ✓.

Problem 6 — Superposition

A node is driven by a 20 V source through 4\ \Omega, a 3 A current source injecting directly, and has 12\ \Omega to ground. Find the node voltage by superposition, then verify by nodal analysis.

Solution

Voltage source acting alone. Kill the current source by opening it. A plain divider:

V' = 20\cdot\frac{12}{4+12} = 15\ \text{V}

Current source acting alone. Kill the voltage source by shorting it, which puts the 4\ \Omega from the node to ground. The 3 A now flows into 4\parallel12 = 3\ \Omega:

V'' = 3\times3 = 9\ \text{V}

Sum:

V = 15+9 = 24\ \text{V}

Verify by nodal:

\frac{V-20}{4}+\frac{V}{12} = 3

Multiply by 12: 3V - 60 + V = 36 \Rightarrow 4V = 96 \Rightarrow V = 24 ✓.

Now the trap. The power in the 12 Ω resistor is 24^2/12 = 48 W. Superposition would wrongly suggest 15^2/12 + 9^2/12 = 18.75 + 6.75 = 25.5 W. Power does not superpose, because it depends on the square. Always find the total voltage or current first, then compute power.

Problem 7 — Thevenin equivalent of a bridge arm

A 12 V source feeds a divider of R_1 = 3\ \text{k}\Omega (top) and R_2 = 6\ \text{k}\Omega (bottom). A second divider from the same source has R_3 = 4\ \text{k}\Omega (top) and R_4 = 4\ \text{k}\Omega (bottom). Find the Thevenin equivalent seen between the two divider midpoints.

Solution

Open-circuit voltage. With nothing bridging them, each divider is independent:

V_A = 12\cdot\frac{6}{3+6} = 8\ \text{V}, \qquad V_B = 12\cdot\frac{4}{4+4} = 6\ \text{V}

V_{Th} = V_A - V_B = 2\ \text{V}

Thevenin resistance. Short the 12 V source. Then at midpoint A, R_1 and R_2 both go to ground, so they are in parallel; same at B:

R_A = 3\text{k}\parallel6\text{k} = 2\ \text{k}\Omega, \qquad R_B = 4\text{k}\parallel4\text{k} = 2\ \text{k}\Omega

Looking between A and B you go through both in series:

R_{Th} = 2+2 = 4\ \text{k}\Omega

Result: 2 V in series with 4 kΩ.

Use it. Bridge a 1\ \text{k}\Omega galvanometer across and the current is 2/(4000+1000) = 0.4 mA. Doing that from the original five-resistor circuit would take nodal analysis; from the equivalent it takes one division.

Note the balance condition. If R_1/R_2 = R_3/R_4 then V_{Th} would be zero and no current would flow whatever you bridged with. Here 3/6 \ne 4/4, so the bridge is unbalanced by 2 V.

Problem 8 — Maximum power transfer

A source has V_{Th} = 18 V and R_{Th} = 6\ \Omega. (a) What load draws maximum power, and how much? (b) How much power reaches a 2 Ω load and a 18 Ω load? (c) What is the efficiency in each case?

Solution

(a) Maximum at R_L = R_{Th} = 6\ \Omega:

I = \frac{18}{12} = 1.5\ \text{A}, \qquad P_L = 1.5^2\times6 = 13.5\ \text{W}

Check against the formula: V_{Th}^2/4R_{Th} = 324/24 = 13.5 W ✓.

(b) With R_L = 2\ \Omega: I = 18/8 = 2.25 A, P_L = 2.25^2\times2 = 10.1 W.

With R_L = 18\ \Omega: I = 18/24 = 0.75 A, P_L = 0.75^2\times18 = 10.1 W.

They are equal. That is not a coincidence — the power curve is symmetric when plotted against \log R_L, so a load three times too big wastes exactly as much as one three times too small.

(c) Efficiency is R_L/(R_{Th}+R_L):

R_L	Power to load	Efficiency
2 Ω	10.1 W	25%
6 Ω	13.5 W	50%
18 Ω	10.1 W	75%

What to notice. The 18 Ω load gets less power but wastes far less. If this were a power supply, 18 Ω is the better operating point; if it were an antenna receiving a faint signal, 6 Ω is the only sensible choice. Maximum power and maximum efficiency are different goals and they never coincide.

Problem 9 — Capacitor energy and the flash

A camera flash uses a 220 µF capacitor charged to 330 V, discharging through a tube in 1.5 ms. Find the stored energy, the average power during the flash, and the charge involved.

Solution

E = \tfrac12CV^2 = 0.5\times220\times10^{-6}\times330^2 = 0.5\times220\times10^{-6}\times108{,}900 = 11.98\ \text{J}

P_{avg} = \frac{E}{t} = \frac{11.98}{1.5\times10^{-3}} = 7987\ \text{W}

Q = CV = 220\times10^{-6}\times330 = 0.0726\ \text{C}

What to notice. Twelve joules is not much energy — a AA battery holds about 10,000 J. But delivering it in 1.5 ms means eight kilowatts of instantaneous power, which is the entire point of the capacitor. A capacitor cannot store much energy, but it can release what it has extremely fast, which is precisely the opposite trade-off from a battery. That single sentence explains where capacitors are used and where batteries are.

Also: 330 V across 0.07 coulombs is genuinely dangerous, and camera flash capacitors have killed people who opened the case after unplugging it. The charge stays for hours.

Problem 10 — RC transient with the universal formula

A 20 V source connects through R_1 = 5\ \text{k}\Omega to a node with R_2 = 20\ \text{k}\Omega to ground and $C = 2.2\ \mu$F to ground. The capacitor is initially charged to 4 V. The switch closes at t = 0. Find v_C(t), the value at 20 ms, and the time to reach 12 V.

Solution

Initial value. Given: v_C(0) = 4 V. Capacitor voltage cannot jump, so this stands at t = 0^+.

Final value. Capacitor becomes an open circuit; the rest is a divider:

v_C(\infty) = 20\cdot\frac{20}{5+20} = 16\ \text{V}

Time constant. Short the source. The capacitor sees 5\text{k}\parallel20\text{k}:

R_{Th} = \frac{5\times20}{25} = 4\ \text{k}\Omega, \qquad \tau = 4000\times2.2\times10^{-6} = 8.8\ \text{ms}

Assemble:

v_C(t) = 16 + (4-16)e^{-t/8.8\text{ms}} = 16 - 12e^{-t/8.8\text{ms}}\ \text{V}

At 20 ms:

v_C = 16 - 12e^{-20/8.8} = 16 - 12e^{-2.273} = 16 - 12(0.1031) = 14.76\ \text{V}

Time to reach 12 V:

12 = 16 - 12e^{-t/\tau} \;\Rightarrow\; e^{-t/\tau} = \frac{4}{12} = 0.3333

-\frac{t}{\tau} = \ln(0.3333) = -1.0986 \;\Rightarrow\; t = 1.0986\times8.8 = 9.67\ \text{ms}

Check at t=0: 16-12 = 4 ✓. Check the trend: it rises from 4 towards 16, and 14.76 at t=2.27\tau is 90% of the way, matching the table ✓.

Problem 11 — Inductor switch-off

A relay coil has L = 250 mH and R = 50\ \Omega, run from 12 V. (a) Find the steady current and the stored energy. (b) If the switch opens and the contacts separate in 2 µs, estimate the peak voltage across the coil. (c) A flyback diode with a 0.7 V forward drop is fitted — now how long does the current take to decay?

Solution

(a) In steady state the inductor is a short circuit as far as DC is concerned, so the current is set by the resistance alone:

I = \frac{12}{50} = 0.24\ \text{A}, \qquad E = \tfrac12LI^2 = 0.5\times0.25\times0.0576 = 7.2\ \text{mJ}

(b) If the current is forced to zero in 2 µs:

v = L\frac{di}{dt} = 0.25\times\frac{0.24}{2\times10^{-6}} = 30{,}000\ \text{V}

In reality the air breaks down at a few kilovolts per millimetre and an arc forms, so you never see 30 kV — you see an arc, contact erosion, and radiated interference that upsets everything nearby. The calculation's job is to show why the arc is inevitable.

(c) With the diode, the current circulates through the coil's own resistance and the diode. The loop equation is

L\frac{di}{dt} + iR + 0.7 = 0

The time constant is \tau = L/R = 0.25/50 = 5 ms, and the current decays roughly exponentially, reaching about 1% after 5\tau = 25 ms.

The trade-off in one line. The diode reduces a 30 kV spike to 0.7 V, but it also makes the relay release about a hundred times more slowly, because the energy now leaks away gently instead of being dumped. Fast-switching designs deliberately use a Zener diode instead, clamping at say 30 V, which is safe for the transistor but drains the energy far quicker.

Problem 12 — RLC damping

A series circuit has L = 10 mH, $C = 1\ \mu$F. (a) Find the natural frequency. (b) What resistance gives critical damping? (c) With R = 20\ \Omega, find the damping ratio, the damped frequency, and describe the response.

Solution

(a)

\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{10^{-2}\times10^{-6}}} = \frac{1}{\sqrt{10^{-8}}} = 10^4\ \text{rad/s}

f_0 = \frac{10^4}{2\pi} = 1592\ \text{Hz}

(b) Critical damping means \zeta = 1:

\zeta = \frac R2\sqrt{\frac CL} = 1 \;\Rightarrow\; R = 2\sqrt{\frac LC} = 2\sqrt{\frac{10^{-2}}{10^{-6}}} = 2\sqrt{10^4} = 200\ \Omega

(c) With R = 20\ \Omega:

\zeta = \frac{20}{2}\sqrt{\frac{10^{-6}}{10^{-2}}} = 10\times10^{-2} = 0.1

Well under 1, so underdamped and ringing.

\omega_d = \omega_0\sqrt{1-\zeta^2} = 10^4\sqrt{1-0.01} = 9950\ \text{rad/s}

which is 1584 Hz — barely different from the undamped 1592 Hz, which is typical: light damping shifts the frequency almost not at all.

The envelope decays as e^{-\zeta\omega_0t} = e^{-1000t}, so the amplitude falls to 1/e in 1 ms, which is about 1.6 cycles. In practice you would see roughly five visible cycles of ringing before it vanished into the noise.

Design note. Q = 1/(2\zeta) = 5 here. The number of visible ring cycles is roughly Q, which is a useful rule of thumb when reading an oscilloscope trace: count the rings and you have estimated the circuit's Q without measuring anything.

Problem 13 — Series RL at AC

A coil of R = 30\ \Omega and L = 80 mH is connected to 230 V, 50 Hz. Find the impedance, current, phase angle, and the real, reactive and apparent power.

Solution

X_L = 2\pi fL = 2\pi(50)(0.08) = 25.13\ \Omega

|\mathbf Z| = \sqrt{30^2+25.13^2} = \sqrt{900+631.5} = \sqrt{1531.5} = 39.1\ \Omega

\theta = \arctan\frac{25.13}{30} = 39.9^\circ \ \text{(current lags)}

I = \frac{230}{39.1} = 5.88\ \text{A}

P = VI\cos\theta = 230\times5.88\times\cos39.9^\circ = 230\times5.88\times0.767 = 1037\ \text{W}

Q = VI\sin\theta = 230\times5.88\times0.641 = 867\ \text{var}

S = VI = 230\times5.88 = 1352\ \text{VA}

Check the triangle: \sqrt{1037^2+867^2} = \sqrt{1{,}075{,}369+751{,}689} = \sqrt{1{,}827{,}058} = 1352 ✓.

Check the real power a second way: all real power is dissipated in the resistance, so P = I^2R = 5.88^2\times30 = 34.6\times30 = 1037 W ✓. That second check is the one to trust.

Problem 14 — Power factor correction

A workshop draws 8 kW at 230 V, 50 Hz with a power factor of 0.72 lagging. Find the present current, then the capacitance needed to correct to 0.95, and the new current.

Solution

Present state.

S = \frac{8000}{0.72} = 11{,}111\ \text{VA}, \qquad I = \frac{11{,}111}{230} = 48.3\ \text{A}

\theta_1 = \arccos(0.72) = 43.9^\circ, \qquad \tan\theta_1 = 0.964

Q_1 = 8000\times0.964 = 7712\ \text{var}

Target.

\theta_2 = \arccos(0.95) = 18.19^\circ, \qquad \tan\theta_2 = 0.329

Q_2 = 8000\times0.329 = 2632\ \text{var}

Capacitor must supply the difference:

Q_C = 7712-2632 = 5080\ \text{var}

C = \frac{Q_C}{\omega V^2} = \frac{5080}{314.16\times230^2} = \frac{5080}{314.16\times52{,}900} = \frac{5080}{16.62\times10^6} = 306\ \mu\text{F}

New current.

S_2 = \frac{8000}{0.95} = 8421\ \text{VA}, \qquad I_2 = \frac{8421}{230} = 36.6\ \text{A}

What was gained. The current dropped from 48.3 A to 36.6 A, a 24% reduction, for the same useful work. Cable losses fall as the square, so they drop by 1-(36.6/48.3)^2 = 43\%. The real power delivered has not changed by one watt — only the useless circulating component has been removed.

Problem 15 — Series resonance in full

A series circuit has R = 15\ \Omega, L = 50 mH, C = 200 nF, driven by 5 V rms. Find f_0, Q, bandwidth, current at resonance, and the voltage across the inductor and capacitor at resonance.

Solution

f_0 = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{0.05\times2\times10^{-7}}} = \frac{1}{2\pi\sqrt{10^{-8}}} = \frac{1}{2\pi\times10^{-4}} = 1592\ \text{Hz}

\omega_0 = 10^4\ \text{rad/s}, \qquad X_L = \omega_0L = 10^4\times0.05 = 500\ \Omega

Check the capacitor matches: X_C = 1/(\omega_0C) = 1/(10^4\times2\times10^{-7}) = 500\ \Omega ✓ — equal and opposite, which is what resonance means.

Q = \frac{X_L}{R} = \frac{500}{15} = 33.3

\text{BW} = \frac{f_0}{Q} = \frac{1592}{33.3} = 47.8\ \text{Hz}

I = \frac VR = \frac{5}{15} = 0.333\ \text{A}

V_L = IX_L = 0.333\times500 = 167\ \text{V}, \qquad V_C = 167\ \text{V (opposite phase)}

Check the magnification: QV = 33.3\times5 = 167 V ✓.

What to notice. A 5 V source produces 167 V across each reactive component. They cancel exactly, so the source only ever sees 5 V across 15 Ω — but if you put a voltmeter across the coil you would read 167 V, and a capacitor rated 100 V would fail. This is the practical trap of resonance, and it is why resonant circuits specify component voltage ratings from QV, not from the supply.

Half-power frequencies are at 1592 \pm 23.9 Hz, so 1568 Hz and 1616 Hz. A tuner this sharp would pass a 48 Hz-wide band — fine for a carrier, far too narrow for FM broadcast audio which needs about 200 kHz.

Problem 16 — A parallel RLC and the tank

An LC tank has L = 2 mH and C = 500 pF, with the coil's winding resistance r = 8\ \Omega in series with the inductor. Find the resonant frequency, the Q, and the effective parallel impedance at resonance.

Solution

f_0 \approx \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{2\times10^{-3}\times5\times10^{-10}}} = \frac{1}{2\pi\sqrt{10^{-12}}} = \frac{1}{2\pi\times10^{-6}} = 159.2\ \text{kHz}

\omega_0 = 10^6\ \text{rad/s}, \qquad X_L = 10^6\times2\times10^{-3} = 2000\ \Omega

Q = \frac{X_L}{r} = \frac{2000}{8} = 250

The parallel impedance at resonance. A coil with series resistance r looks, at resonance, like a pure resistance of

R_p = Q^2r = 250^2\times8 = 500{,}000\ \Omega = 500\ \text{k}\Omega

Equivalently R_p = QX_L = 250\times2000 = 500 kΩ ✓, and also R_p = L/(rC) = 0.002/(8\times5\times10^{-10}) = 500 kΩ ✓ — three routes, same answer.

What this means. Eight ohms of copper resistance, viewed through a resonance with Q = 250, presents itself as half a megohm. The tank is nearly invisible to the source at its resonant frequency and a near-short at every other frequency, which is exactly the behaviour a radio front end wants: enormous response at the wanted frequency, nothing anywhere else.

Bandwidth: 159.2\ \text{kHz}/250 = 637 Hz. That is far too sharp for a real receiver, which is why practical tank circuits are deliberately loaded with a resistor to lower Q and widen the passband to whatever the signal actually needs.

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Part 2 leaves linear components behind. Everything from here has a curve rather than a straight line, and the curve is where amplification comes from.