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 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 I3 in branch 3, flowing up into node A in the direction of the arrow.
The current source is adjusted until the node voltage at A is VA = 11.5 V. Use nodal analysis to find the source current Is (direction as shown).
Find the Thévenin voltage Vth seen from terminals a–b.
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 voltage vL across the inductor at t = 5 ms.
Just before the switch closes at t = 0, the capacitor holds 20 V with the polarity shown. How long does it take for vC to reach 17 V? Find t.
A series RLC circuit is connected to a 120 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 15 kW at a power factor of 0.65 lagging from a 400 V supply. Find the line current I.
A 75 kVA single-phase transformer supplies a load of 21 kW at a power factor of 0.65 lagging. How much extra real power ΔP can be added as a resistive (unity power factor) load before the transformer reaches its kVA rating?
A balanced Y-connected load with an impedance of 12 + j5 Ω per phase is connected to a balanced three-phase supply with a line-to-line voltage of 208 V. Find the total real power P absorbed by the load in kW.
A balanced three-phase load takes 60 kW at a power factor of 0.65 lagging from a 690 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.
A single-phase 15 kVA, 480/240 V transformer operates at its rated load. Treat it as ideal. Find the current ILV in the low-voltage winding.
A single-phase 250 kVA transformer has a core loss of 2000 W. At rated load and a power factor of 0.85 lagging its efficiency is 97%. Find the full-load copper loss Pcu,fl in watts.
Answer: Req = 25.45 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: Req = 14.21 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: I3 = −3.02 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: Is = 0.5167 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.
VA is known, so this equation gives VB.
Currents leaving B through the resistors equal the current pushed into B by sources.
Answer: Vth = 10 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, no current flows through R₃, so there is no drop across it and V_b = V_A. R₁ and R₂ form a voltage divider.
Answer: IN = 2.232 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: vL = 3.994 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.
Long after the switch closes the inductor behaves like a short circuit, so the final current is set by R alone.
At t = 0 the whole supply voltage appears across the inductor; it then decays to zero.
Answer: t = 117.7 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 = 0.399 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: VL = 4.126 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 = 2.063 times the supply voltage.
Answer: I = 57.69 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: ΔP = 49.87 kW
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.
At full rating S = Smax, and Q is unchanged by the resistive load.
Answer: P = 3.072 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.
Only the resistive part of each impedance absorbs real power.
Check: cos φ = R/|Z| = 0.9231, and √3·VL·IL·cos φ gives the same power.
Answer: QC = 57.96 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: ILV = 62.5 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.
Both windings carry the same apparent power.
Answer: Pcu,fl = 4572 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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