Formula sheets, the key method steps for every topic and 16 practice problems, with full solutions at the back.
Find the equivalent resistance Req seen by the source V1.
Find the current IR2 through R2.
Use nodal analysis to find the current IR3 through R3 in the direction of the arrow.
Use nodal analysis to find the node voltage VB (measured from ground).
A load RL = 3 Ω is connected across a–b. Replace the rest of the circuit with its Thévenin equivalent and find the load current IL.
Find the Norton current IN: the current that flows from a to b if the terminals are shorted.
The inductor carries no current before the switch closes at t = 0. Find the energy W stored in the inductor at t = 1 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 capacitor voltage vC at t = 80 ms.
A series RLC circuit is connected to a 12 V (rms), 100 Hz supply. Find the phase angle φ of the impedance, in degrees (positive when the current lags the voltage).
A resistor R1 is connected in series with a parallel combination of R2 and a capacitor C, across a 120 V (rms), 200 Hz supply. Find the rms current I.
A single-phase motor takes 32 kW at a power factor of 0.75 lagging from a 400 V supply. Find the line current I.
A 230 V (rms), 60 Hz source has an internal impedance Zs = 2 + j10 Ω. The load impedance ZL can be chosen freely. The load for maximum power is built from a resistor in series with a capacitor. Find the capacitance C.
A three-phase induction motor delivers 110 kW of mechanical output power at full load. At this load its efficiency is 91% and its power factor is 0.82 lagging. The supply has a line-to-line voltage of 690 V. Find the line current IL.
A balanced three-phase load takes 60 kW at a power factor of 0.72 lagging from a 400 V (line-to-line), 50 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.9 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 = 4:1 has 480 V applied to its primary winding. A 60 Ω resistor is connected across the secondary winding. Find the primary current I1.
A single-phase 25 kVA transformer has a core loss of 75 W and a copper loss of 187.5 W at rated load. It supplies 25% of its rated kVA at a power factor of 0.9 lagging. Find the total losses Ploss in watts at this load.
Answer: Req = 14.29 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: IR2 = 0.2791 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.
Subtract the drop across R₁ to get the voltage at node A, which is the voltage across R₂.
Answer: IR3 = −0.6383 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). 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: VB = −1.235 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.
Ground is the bottom wire. Known node voltages come straight from the sources (V1 = 9 V, V2 = −9 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 = 0.9146 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: IN = 2.611 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.
A Norton source is the same circuit seen as a current source in parallel with R_th. Its current is the short-circuit current at a–b.
Answer: W = 7.992 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: vC = 5.698 V
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: φ = −65.87 °
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).
Answer: I = 0.741 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|.
Multiply top and bottom by the complex conjugate to split Zp into real and imaginary parts.
Answer: I = 106.7 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: C = 265.3 µF
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 = 123.3 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 = 28.77 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 = 0.5 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 = 960 Ω, and 480/960 gives the same current.
Answer: Ploss = 86.72 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 = 98.48%.
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