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.
The diamond is a voltage-controlled current source that pushes a current of 0.1·VA (in A, with VA in V) up into node B. Use nodal analysis to find the current IR4 through R4 in the direction of the arrow.
A load RL = 20 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. How long after the switch closes does the current reach 80% of its final value? Find t.
Just before the switch closes at t = 0, the capacitor holds 24 V with the polarity shown. How long does it take for vC to reach 21.2 V? Find t.
A series RLC circuit with L = 2 mH must resonate at 2000 Hz. Find the capacitance C needed.
A series RLC circuit is connected to a 230 V (rms), 60 Hz supply. Find the resonant frequency f0 of the circuit.
Two loads are connected in parallel to a 240 V single-phase supply: a 29 kW load at a power factor of 0.85 lagging and a 6 kW load at a power factor of 0.85 leading. Find the overall power factor cos φ seen by the supply.
A single-phase load takes 35 kW at a power factor of 0.7 lagging from a 400 V, 50 Hz supply. A capacitor connected in parallel raises the power factor to 0.9. Find the capacitance C needed.
A three-phase induction motor delivers 55 kW of mechanical output power at full load. At this load its efficiency is 91% and its power factor is 0.9 lagging. The supply has a line-to-line voltage of 400 V. Find the line current IL.
A balanced three-phase load takes 100 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. Find the total reactive power QC the capacitor bank must supply, in kvar.
An 8 Ω resistor is connected across the secondary winding of an ideal transformer with a turns ratio N₁:N₂ = 6:1. Find the resistance R1′ seen looking into the primary winding.
A single-phase 200 kVA transformer has a core loss of 1000 W. At rated load and a power factor of 0.85 lagging its efficiency is 98.2%. Find the full-load copper loss Pcu,fl in watts.
Answer: Req = 9.852 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 = 0.44 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 = −0.9032 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 = 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: IR4 = 0.1975 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.
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 = 0.03922 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 = 0.561 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: t = 0.7082 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.
Long after the switch closes the inductor behaves like a short circuit, so the final current is set by R alone.
Answer: t = 56.59 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: C = 3.166 µF
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 inductive and capacitive reactances cancel: ωL = 1/(ωC).
Answer: f0 = 50.33 Hz
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 inductive and capacitive reactances cancel: ωL = 1/(ωC). The supply frequency does not matter.
Answer: cos φ = 0.9261
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.
A lagging load takes reactive power (Q > 0); a leading load supplies it (Q < 0).
Answer: C = 373.1 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 real power P does not change; only the reactive power drawn from the supply does.
A capacitor across the supply delivers Q = V²·ωC.
Answer: IL = 96.93 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 = 37.63 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: R1′ = 288 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: Pcu,fl = 2116 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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