Formula sheets, the key method steps for every topic and 16 practice problems, with full solutions at the back.
Find the current IR1 delivered by the source, which flows through R1.
Find the equivalent resistance Req seen by the source V1.
The voltage source Vₛ between A and B has its + terminal at B, so it makes VB − VA = Vₛ. Use nodal analysis with a supernode to find the node voltage VB (measured from ground).
The diamond is a voltage-controlled current source that pushes a current of 0.25·VA (in A, with VA in V) up into node B. Use nodal analysis to find the node voltage VA (measured from ground).
A load RL = 12 Ω is connected across a–b. Replace the rest of the circuit with its Thévenin equivalent and find the load current IL.
Find the Thévenin voltage Vth seen from terminals a–b.
The inductor carries no current before the switch closes at t = 0. Find the current i at t = 5 ms.
Just before the switch closes at t = 0, the capacitor holds 6 V with the polarity shown. How long does it take for vC to reach 11.4 V? Find t.
A series RLC circuit is connected to a 12 V (rms), 100 Hz supply. Find the rms current I.
A series RLC circuit is connected to a 50 V (rms), 500 Hz supply. Find the rms voltage VC across the capacitor.
A load with impedance Z = 20 + j15 Ω is connected to a 230 V (rms) single-phase supply. Find the apparent power S.
A single-phase load takes 15 kW at a power factor of 0.75 lagging from a 240 V, 60 Hz supply. A 100 µF capacitor is connected in parallel with the load. Find the new line current I.
A three-phase induction motor delivers 55 kW of mechanical output power at full load. At this load its efficiency is 92% and its power factor is 0.8 lagging. The supply has a line-to-line voltage of 415 V. Find the line current IL.
A balanced three-phase load takes 250 kW at a power factor of 0.65 lagging from a 480 V (line-to-line), 60 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.98 lagging. The real power stays the same. Find the total reactive power QC the capacitor bank must supply, in kvar.
An ideal transformer with a turns ratio N₁:N₂ = 20:1 has 4600 V applied to its primary winding. A 5 Ω resistor is connected across the secondary winding. Find the primary current I1.
A single-phase 100 kVA transformer has a core loss of 600 W and a copper loss of 1800 W at rated load. It supplies 25% of its rated kVA at a power factor of 0.85 lagging. Find the total losses Ploss in watts at this load.
Answer: IR1 = 1.597 A
Series and parallel resistors: in series the same current flows through each resistor and R = R₁ + R₂ + …. In parallel each resistor has the same voltage and 1/R = 1/R₁ + 1/R₂ + …. A ladder network is reduced one pair at a time until a single equivalent resistance remains.
Start at the end farthest from the source and work back towards it. Resistors carrying the same current are in series; resistors across the same two nodes are in parallel.
The source sees only Req, so Ohm’s law gives the current it delivers. That current flows through R₁.
Answer: Req = 12.02 Ω
Series and parallel resistors: in series the same current flows through each resistor and R = R₁ + R₂ + …. In parallel each resistor has the same voltage and 1/R = 1/R₁ + 1/R₂ + …. A ladder network is reduced one pair at a time until a single equivalent resistance remains.
Start at the end farthest from the source and work back towards it. Resistors carrying the same current are in series; resistors across the same two nodes are in parallel.
Answer: VB = 6.8 V
Nodal analysis: choose a reference node, label the unknown node voltages, write Kirchhoff’s current law at each unknown node using Ohm’s law (I = ΔV/R), and solve the resulting equations.
The voltage source between A and B carries an unknown current, so KCL cannot be written at A or B separately. Enclose A, B and the source in one supernode: the current entering it equals the current leaving it.
Answer: VA = 7.529 V
Nodal analysis: choose a reference node, label the unknown node voltages, write Kirchhoff’s current law at each unknown node using Ohm’s law (I = ΔV/R), and solve the resulting equations.
Currents leaving A through the resistors equal the current pushed into A by sources.
The dependent source pushes g·VA into node B. Treat it like an independent source, then move the VA term to the left-hand side.
Answer: IL = 0.5854 A
Thévenin’s theorem: any linear two-terminal network behaves like a single voltage source Vth in series with a resistance Rth. Norton’s theorem gives the same network as a current source IN = Vth/Rth in parallel with Rth.
Replace voltage sources with shorts and current sources with opens, then look into the circuit from a–b.
With a–b open, no current flows through R₃, so there is no drop across it and V_b = V_A. R₁ and R₂ form a voltage divider.
The Thévenin source and R_th in series behave exactly like the original circuit at a–b, so the load sees a simple series loop.
Answer: Vth = 33.5 V
Thévenin’s theorem: any linear two-terminal network behaves like a single voltage source Vth in series with a resistance Rth. Norton’s theorem gives the same network as a current source IN = Vth/Rth in parallel with Rth.
Replace voltage sources with shorts and current sources with opens, then look into the circuit from a–b.
With a–b open, the only path for Iₛ is through R₃ into node A. Find V_A by superposition of V₁ and Iₛ, then add the drop across R₃.
Answer: i = 0.2203 A
First-order transients: in a circuit with one capacitor or one inductor, every voltage and current moves exponentially from its initial value to its final value: x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ, with τ = RC or τ = L/R. After about 5τ the circuit has settled.
Long after the switch closes the inductor behaves like a short circuit, so the final current is set by R alone.
Answer: t = 99.78 ms
First-order transients: in a circuit with one capacitor or one inductor, every voltage and current moves exponentially from its initial value to its final value: x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ, with τ = RC or τ = L/R. After about 5τ the circuit has settled.
The capacitor voltage cannot jump, so it starts at its stored value and heads for the source voltage.
Answer: I = 0.2634 A
AC circuits with phasors: resistors, inductors and capacitors have impedances R, jωL and 1/(jωC), with ω = 2πf. In series, impedances add like resistances, but as complex numbers. Rms current and voltage follow Ohm’s law with |Z|.
In series the impedances add as complex numbers: Z = R + j(XL − XC).
The same current flows through every element of a series circuit.
Answer: VC = 8.146 V
AC circuits with phasors: resistors, inductors and capacitors have impedances R, jωL and 1/(jωC), with ω = 2πf. In series, impedances add like resistances, but as complex numbers. Rms current and voltage follow Ohm’s law with |Z|.
In series the impedances add as complex numbers: Z = R + j(XL − XC).
The same current flows through every element of a series circuit.
Answer: S = 2116 VA
AC power: apparent power S = V·I (VA) splits into real power P = S·cos φ (W), which does useful work, and reactive power Q = S·sin φ (var), which moves back and forth between source and load. The power factor is cos φ. A capacitor in parallel supplies reactive power locally, so the line current drops for the same real power.
Answer: I = 77.65 A
AC power: apparent power S = V·I (VA) splits into real power P = S·cos φ (W), which does useful work, and reactive power Q = S·sin φ (var), which moves back and forth between source and load. The power factor is cos φ. A capacitor in parallel supplies reactive power locally, so the line current drops for the same real power.
The real power P does not change; only the reactive power drawn from the supply does.
Answer: IL = 104 A
Balanced three-phase systems: in a Y (star) connection the line-to-line voltage is √3 times the phase voltage, VL = √3·Vph, and the line current equals the phase current. In a Δ (delta) connection the line-to-line voltage equals the phase voltage and the line current is √3 times the phase current, IL = √3·Iph. The total power is three times the power of one phase: P = 3·Vph·Iph·cos φ = √3·VL·IL·cos φ.
Answer: QC = 241.5 kvar
Balanced three-phase systems: in a Y (star) connection the line-to-line voltage is √3 times the phase voltage, VL = √3·Vph, and the line current equals the phase current. In a Δ (delta) connection the line-to-line voltage equals the phase voltage and the line current is √3 times the phase current, IL = √3·Iph. The total power is three times the power of one phase: P = 3·Vph·Iph·cos φ = √3·VL·IL·cos φ. Power factor correction: capacitors supply reactive power locally, QC = P·(tan φ1 − tan φ2). Each of the three capacitors supplies QC/3 = V²·ω·C, where V is the voltage across that capacitor: VL in a Δ bank, VL/√3 in a Y bank.
The real power is unchanged, so each power factor fixes the reactive power: Q = P·tan φ.
Answer: I1 = 2.3 A
Ideal transformer: the voltages are in the ratio of the turns and the currents in the inverse ratio: V₁/V₂ = N₁/N₂ = a and I₁/I₂ = N₂/N₁ = 1/a, so the power in equals the power out, V₁·I₁ = V₂·I₂. A load impedance Z on the secondary looks like a²·Z from the primary.
An ideal transformer has no losses: V₁·I₁ = V₂·I₂.
Check: the load seen from the primary is a²·RL = 2000 Ω, and 4600/2000 gives the same current.
Answer: Ploss = 712.5 W
Transformer losses: the core loss (hysteresis and eddy currents) depends on the voltage and is the same at any load. The copper loss in the windings goes with the current squared, so at a fraction x of rated load it is x²·Pcu,fl. Efficiency η = Pout/(Pout + Pcore + Pcu), with Pout = x·S·pf. The efficiency is highest at the load where the copper loss equals the core loss.
The core loss is the same at any load.
The efficiency at this load is η = 96.76%.
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