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.
Three branches are connected between node A and ground, as shown. Use nodal analysis to find the current I1 in branch 1, flowing up into node A in the direction of the arrow.
Use nodal analysis to find the current IR1 through R1 in the direction of the arrow.
A load RL = 4 Ω is connected across a–b. Replace the rest of the circuit with its Thévenin equivalent and find the load current IL.
A load RL = 5 Ω is connected across a–b. Replace the rest of the circuit with its Thévenin equivalent and find the load current IL.
The inductor carries no current before the switch closes at t = 0. Find the energy W stored in the inductor at t = 5 ms, in mJ.
The switch has been closed for a long time, so the circuit is in steady state. At t = 0 the switch opens. Find the current i2 through R2 at t = 30 ms.
A resistor and a capacitor are connected in parallel across a 12 V (rms), 500 Hz supply. Find the rms current I.
A series RLC circuit is driven by a 2 V (rms) source whose frequency is set to the resonant frequency of the circuit. Find the rms voltage VL across the inductor.
A single-phase motor takes 32 kW at a power factor of 0.65 lagging from a 240 V supply. Find the line current I.
A 24 V (rms), 60 Hz source has an internal impedance Zs = 4 + j15 Ω. The load impedance ZL can be chosen freely. Find the maximum real power Pmax that can be delivered to the load.
A three-phase induction motor delivers 45 kW of mechanical output power at full load. At this load its efficiency is 95% and its power factor is 0.9 lagging. The supply has a line-to-line voltage of 415 V. Find the line current IL.
A balanced three-phase load takes 60 kW at a power factor of 0.8 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 amplifier with an output resistance of 675 Ω drives a 75 Ω loudspeaker through an ideal transformer. For maximum power transfer the loudspeaker must look like 675 Ω from the amplifier. Find the turns ratio a = N₁/N₂.
A single-phase 200 kVA transformer has a core loss of 1200 W. At rated load and a power factor of 0.9 lagging its efficiency is 98.5%. Find the full-load copper loss Pcu,fl in watts.
Answer: IR1 = 0.375 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 = 16.77 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: I1 = 1.515 A
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.
Take the bottom wire as ground. Each branch carries (VA − Vk)/Rk away from A, where Vk is the voltage of its source (zero for a plain resistor). The currents leaving A add up to zero.
The arrow points up through the branch, into node A.
A negative result means the real current flows against the arrow.
Answer: IR1 = 1.565 A
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.
Ground is the bottom wire. Known node voltages come straight from the sources (V1 = 10 V, V2 = −18 V). Unknowns: VA, VB.
Currents leaving A through the resistors equal the current pushed into A by sources.
Currents leaving B through the resistors equal the current pushed into B by sources.
Answer: IL = 0.1463 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: IL = 1.184 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, 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.
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: W = 4.995 mJ
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: i2 = 1.754 mA
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.
In steady state the capacitor is an open circuit, so R₁ and R₂ form a voltage divider. The capacitor voltage cannot jump when the switch opens.
Answer: I = 0.5117 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|.
Both branches are connected straight across the supply, so each sees the full voltage V.
The resistor current is in phase with V and the reactive current is 90° out of phase, so they add like the sides of a right triangle.
Answer: VL = 28.28 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|.
At resonance the reactances cancel, so the impedance is just R.
This is Q = 14.14 times the supply voltage.
Answer: I = 205.1 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.
Only the real part of the power does work, so divide by the power factor.
Answer: Pmax = 36 W
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 load reactance cancels the source reactance, and the load resistance equals the source resistance.
Answer: IL = 73.22 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 = 32.82 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: a = 3
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.
Answer: Pcu,fl = 1541 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.
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