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 power PIs delivered by the current source Is.
Use nodal analysis to find the node voltage VA (measured from ground).
Find the Norton current IN: the current that flows from a to b if the terminals are shorted.
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.
The source Vₛ charges the capacitor through R₁, and R₂ is connected in parallel with C. The capacitor is uncharged when the switch closes at t = 0. Find the time constant τ of the circuit after the switch closes.
A timer starts charging an uncharged capacitor through a resistor from a 9 V supply at t = 0. It must trigger when the capacitor voltage reaches 7.2 V, exactly 250 ms after the start. With C = 1 µF, find the resistance R needed.
A series RLC circuit is connected to a 50 V (rms), 200 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 230 V (rms), 500 Hz supply. Find the rms current I.
A single-phase motor takes 31 kW at a power factor of 0.75 lagging from a 230 V supply. Find the reactive power Q it draws.
A single-phase load takes 25 kW at a power factor of 0.75 lagging from a 400 V, 60 Hz supply. A capacitor connected in parallel raises the power factor to 0.92. Find the new line current Inew.
A balanced three-phase load draws a line current of 150 A from a supply with a line-to-line voltage of 415 V. The power factor is 0.8 lagging. Find the total real power P drawn by the load in kW.
A balanced three-phase load takes 250 kW at a power factor of 0.78 lagging from a 415 V (line-to-line), 50 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.92 lagging. The real power stays the same. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.
A 4 Ω resistor is connected across the secondary winding of an ideal transformer with a turns ratio N₁:N₂ = 12:1. Find the resistance R1′ seen looking into the primary winding.
A single-phase 75 kVA transformer has a core loss of 600 W and a copper loss of 1800 W at rated load. At what load Sη,max, in kVA, is its efficiency highest?
Answer: Req = 16.43 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 = 0.3052 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: PIs = 6.923 W
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.
Currents leaving B through the resistors equal the current pushed into B by sources.
The source pushes Iₛ up into node B, from ground (0 V) to VB. A source whose current leaves its + end delivers P = V·I.
Answer: VA = 15.35 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 = 36 V, V2 = −12 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: IN = 0.383 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.
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: IL = 1.257 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: τ = 545.9 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.
Short the source to find Rth; open the capacitor to find the voltage it will settle at.
Answer: R = 155.3 kΩ
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.
kΩ times µF gives milliseconds.
Answer: φ = −79.36 °
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 = 6.23 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: Q = 27.34 kvar
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: Inew = 67.93 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: P = 86.26 kW
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: Cph = 579.5 µF
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.
In a Δ bank each capacitor is connected between two lines.
Answer: R1′ = 576 Ω
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.
The voltage goes up by a and the current down by a, so the ratio V/I goes up by a².
Answer: Sη,max = 43.3 kVA
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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