A set of fully worked problems on charge, current, voltage, power, energy, resistance and resistivity, and the average / RMS values of time-varying waveforms. Each problem is followed by a complete, step-by-step solution with the final result highlighted.
EE Core · Electric Circuit Analysis · 20 solved problems
Demonstrative VideoWalkthrough
Problem 1Charge & Current
The figure shows the current flowing through a capacitor. Determine the charge acquired by the capacitor during the first \(5\,\mu\text{s}\).
Current through the capacitor as a function of time.
Solution
Current and charge are related by i = dq/dt, so the charge is the area under the \(i\!-\!t\) curve. Over the first \(5\,\mu\text{s}\), add the areas of the elementary triangles and rectangles under the graph:
Decomposition of the area into triangles and rectangles.
AnswerCharge acquired \(Q = 15\ \mu\text{C}\)
Problem 2Parallel Resistors / Power
Two coils connected in parallel across a 100 V DC supply draw 10 A from the supply. The power dissipated in one coil is 600 W. Find the resistance of each coil.
Solution
The effective (parallel) resistance seen by the supply:
Number of 242 Ω lamps in series to obtain 484 Ω: \(n = 484 / 242 = 2\). Check the power drawn at 220 V:
\[ P = \frac{220^2}{484} = 100\ \text{W} \checkmark \]
Two 200 W lamps in series give an equivalent of 484 Ω.
AnswerTwo lamps in series (\(n = 2\))
Problem 4Sources & Power
When a resistor \(R\) is connected to a current source it dissipates 18 W. When the same \(R\) is connected to a voltage source whose numerical magnitude equals that of the current source, it absorbs 4.5 W. Find the magnitude of the source and the value of \(R\).
Solution
With a current source of value \(I\) driving \(R\):
\[ P_1 = I^2 R = 18\ \text{W} \]
The voltage source has the same magnitude, i.e. \(V = I\) numerically, so:
The charge \(q(t)\) delivered by a constant-voltage source is shown. Determine the current supplied by the source at (a) \(t = 1\,\text{s}\) and (b) \(t = 3\,\text{s}\).
Solution
The current is the slope of the charge waveform, i = dq/dt.
aSegment A→B, valid for \(0 \le t \le 2\,\text{s}\):
\[ I = \frac{V}{R} = \frac{220}{57.6} = 3.82\ \text{A} \]
Since \(3.82\,\text{A} < 4.167\,\text{A}\), the element is not damaged. Power consumed:
\[ P' = V I = 220 \times 3.82 = 840.4\ \text{W} \]
AnswerNot damaged; output falls to \(\approx 840\ \text{W}\) (rating reduced)
Problem 7Power Rating / Series
What is the maximum voltage that can be applied across the series combination of a 150 Ω, 2 W resistor and a 100 Ω, 1 W resistor without exceeding the power rating of either resistor?
Solution
From \(P = I^2 R\), the maximum safe current of each resistor is \(I_{\max} = \sqrt{P/R}\). For the 150 Ω resistor:
A wire 50 m long and 2 mm² in cross-section has a resistance of 0.56 Ω. A 100 m length of the same material has a resistance of 2 Ω at the same temperature. Find the diameter of this second wire.
Solution
Both wires share the same material, so the resistivities are equal, \(\rho_1 = \rho_2\). Given:
The voltage across a 10 Ω resistor is the square wave shown (amplitude \(\pm 10\,\text{V}\), period \(\pi\)). Find the average power dissipated by the resistor.
Solution
Average power is \(P_{\text{avg}} = V_{\text{rms}}^2/R\), where \(V_{\text{rms}}^2 = \tfrac{1}{T}\int_0^T v^2\,dt\). Because power depends on \(v^2\), the sign of the voltage is irrelevant — both half-cycles contribute positively:
The squared voltage is constant at 100 V² over the whole period.
Answer\(P_{\text{avg}} = 10\ \text{W}\)
Problem 10Resistivity / Geometry
A wire-wound resistor is to be made from 0.2 mm diameter constantan wire wound around a cylinder of 1 cm diameter. How many turns are needed for a resistance of 50 Ω? Take the resistivity of constantan as \(49 \times 10^{-8}\ \Omega\text{m}\).
Answer\(I \approx 10.5\ \text{A};\quad E = 14.478\ \text{kWh/day}\)
Problem 13Inductor Energy
The figure shows the current through a practical inductor of resistance 1 Ω and inductance 2 H. The current ramps linearly from 0 to 6 A during the first 2 s and then stays at 6 A. Find the total energy absorbed by the inductor in the first four seconds.
Solution
Energy stored in the magnetic field of the 2 H inductance between \(t=0\) and \(t=4\,\text{s}\) (here \(i(0)=0,\ i(4)=6\,\text{A}\)):
The current entering the positive terminal of a device is \(i(t) = 3e^{-2t}\ \text{A}\) and the voltage across it is \(v(t) = 5\,\dfrac{di}{dt}\ \text{V}\).
Find the charge delivered between \(t = 0\) and \(t = 2\,\text{s}\).
Interpretation: the negative power and energy mean the device is actually delivering energy to the rest of the circuit, not absorbing it.
Answer\(q = 1.4725\ \text{C},\quad p = -90e^{-4t}\ \text{W},\quad W = -22.5\ \text{J}\)
Problem 15Power / Energy (AC)
The charge entering the positive terminal of an element is \(q = 10\sin 4\pi t\ \text{mC}\), while the voltage across it (plus to minus) is \(v = 2\cos 4\pi t\ \text{V}\).
Find the power delivered to the element at \(t = 0.3\,\text{s}\).
Calculate the energy delivered between 0 and 0.6 s.
Solution
Current:
\[ i = \frac{dq}{dt} = 40\pi\cos 4\pi t\ \text{mA} \]
aPower:
\[
\begin{aligned}
p &= vi = 80\pi\cos^2 4\pi t\ \text{mW}\\[2pt]
p(0.3) &= 80\pi\cos^2(4\pi \times 0.3) = 164.5\ \text{mW}
\end{aligned}
\]
bEnergy, using \(\cos^2\theta = \tfrac{1}{2}(1+\cos 2\theta)\):
Answer\(p(0.3\,\text{s}) = 164.5\ \text{mW},\quad W = 78.34\ \text{mJ}\)
Additional Practice Problems
The following problems extend the same fundamentals — temperature dependence of resistance, RMS values, series–parallel reduction, energy billing and capacitor energy — at a matching difficulty level, each with a complete solution.
Problem 16Temperature & Resistance
A copper field coil has a resistance of 50 Ω at 20 °C. The temperature coefficient of resistance of copper at 20 °C is \(\alpha_{20} = 0.00393\ /^{\circ}\text{C}\). Find its resistance when the coil heats up to 70 °C in service.
Solution
Resistance varies linearly with temperature as \(R_T = R_0\left[1 + \alpha_0\,(T - T_0)\right]\):
Takeaway: a 50 °C rise increases copper resistance by about 20 %, which is why winding resistance (and \(I^2R\) loss) must be specified at the operating temperature, not at ambient.
Answer\(R_{70} \approx 59.83\ \Omega\)
Problem 17Average & RMS Values
A current rises linearly from 0 to 10 A over 2 s, drops instantly to 0, and repeats (a sawtooth of period 2 s). Find (a) its average value, (b) its RMS value, and (c) the power it delivers to a 4 Ω resistor.
Solution
Over one period the current is \(i(t) = 5t\) for \(0 \le t \le 2\,\text{s}\).
A 12 V battery is connected to a 6 Ω resistor in series with the parallel combination of a 4 Ω and a 12 Ω resistor. Find (a) the total current drawn from the battery and (b) the power dissipated in the 4 Ω resistor.
A household runs a 2 kW air-conditioner for 8 hours a day and ten 15 W LED bulbs for 6 hours a day. If electricity costs ₹7.50 per kWh, estimate the monthly (30-day) energy bill.
Solution
Daily energy of each load (energy = power × time):
A 100 μF capacitor is charged to 200 V. (a) Find the charge and energy stored. (b) This capacitor is then connected across an identical, initially uncharged 100 μF capacitor. Find the final common voltage and the energy lost in the process.
Key idea: charge sharing always loses energy — here exactly half — regardless of how small the connecting resistance is. The lost energy is independent of \(R\).
