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 current IR1 delivered by the source, which flows through R1.
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.
Find the Thévenin resistance Rth seen from terminals a–b.
A load RL = 30 Ω is connected across a–b. Use a Thévenin equivalent to find the power PL delivered to the load.
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.
A timer starts charging an uncharged capacitor through a resistor from a 15 V supply at t = 0. It must trigger when the capacitor voltage reaches 9.48 V, exactly 10 ms after the start. With R = 1 kΩ, find the capacitance C needed.
A series RLC circuit with C = 47 µF must resonate at 5000 Hz. Find the inductance L needed.
A resistor R1 is connected in series with a parallel combination of R2 and a capacitor C, across a 50 V (rms), 200 Hz supply. Find the magnitude of the impedance |Z|.
A single-phase motor takes 9 kW at a power factor of 0.7 lagging from a 240 V supply. Find the line current I.
A single-phase load takes 45 kW at a power factor of 0.8 lagging from a 240 V, 60 Hz supply. A capacitor connected in parallel raises the power factor to 0.95. Find the new line current Inew.
A balanced three-phase load draws a line current of 12 A from a supply with a line-to-line voltage of 690 V. The power factor is 0.95 lagging. Find the total real power P drawn by the load in kW.
A balanced three-phase load takes 200 kW at a power factor of 0.75 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 0.9 lagging. The real power stays the same. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.
An ideal transformer with a turns ratio N₁:N₂ = 10:1 has 2400 V applied to its primary winding. A 40 Ω resistor is connected across the secondary winding. Find the primary current I1.
A single-phase 25 kVA transformer has a core loss of 150 W and a copper loss of 450 W at rated load. At what load Sη,max, in kVA, is its efficiency highest?
Answer: IR1 = 0.8149 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: IR1 = 1.664 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: IR3 = −0.6842 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 = 6 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 = 0.2979 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: Rth = 11.6 Ω
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.
Answer: PL = 1.771 W
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 = 16.09 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: C = 10 µF
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: L = 0.02156 mH
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: |Z| = 85.42 Ω
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 = 53.57 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: Inew = 197.4 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 = 13.62 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 = 489.9 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: I1 = 0.6 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 = 4000 Ω, and 2400/4000 gives the same current.
Answer: Sη,max = 14.43 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.
ExamProblems · examproblems.com/courses/electrical-engineering-fundamentals