Formula sheets, the key method steps for every topic and 16 practice problems, with full solutions at the back.
Find the current IR2 through R2.
Find the equivalent resistance Req seen by the source V1.
The voltage source Vₛ between A and B has its + terminal at B, so it makes VB − VA = Vₛ. Use nodal analysis with a supernode to find the current Ix through the voltage source, from A to B as shown by the arrow.
The current source is adjusted until the node voltage at A is VA = 9 V. Use nodal analysis to find the source current Is (direction as shown).
Find the Norton current IN: the current that flows from a to b if the terminals are shorted.
Find the Norton current IN: the current that flows from a to b if the terminals are shorted.
The inductor carries a steady current of 3 A. At t = 0 the switch closes the loop through R and the source is removed, so the inductor current decays through R. How long does it take for the current to fall to 0.75 A? Find t.
The switch across R2 has been open for a long time, so the inductor current is steady. At t = 0 the switch closes and shorts out R2. Find the inductor current i at t = 10 ms.
A series RLC circuit is connected to a 100 V (rms), 60 Hz supply. Find the phase angle φ of the impedance, in degrees (positive when the current lags the voltage).
A series RLC circuit is connected to a 100 V (rms), 200 Hz supply. Find the rms current I.
Two loads are connected in parallel to a 400 V single-phase supply: a 10 kW load at a power factor of 0.7 lagging and a 15 kW load at a power factor of 0.9 leading. Find the overall power factor cos φ seen by the supply.
A single-phase load takes 30 kW at a power factor of 0.8 lagging from a 400 V, 60 Hz supply. A capacitor connected in parallel raises the power factor to 0.98. Find the capacitance C needed.
A three-phase induction motor delivers 45 kW of mechanical output power at full load. At this load its efficiency is 93% 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 200 kW at a power factor of 0.8 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 0.98 lagging. The real power stays the same. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.
An amplifier with an output resistance of 128 Ω drives a 2 Ω loudspeaker through an ideal transformer. For maximum power transfer the loudspeaker must look like 128 Ω from the amplifier. Find the turns ratio a = N₁/N₂.
A single-phase 200 kVA transformer has a core loss of 1600 W and a copper loss of 2400 W at rated load. At what load Sη,max, in kVA, is its efficiency highest?
Answer: IR2 = 0.4909 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: Req = 16.38 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: Ix = 1.714 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.
The voltage source between A and B carries an unknown current, so KCL cannot be written at A or B separately. Enclose A, B and the source in one supernode: the current entering it equals the current leaving it.
Everything that passes through the source from A to B leaves node B through R₂.
A negative result means the real current flows against the arrow.
Answer: Is = 2.75 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: IN = 0.7895 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: IN = 4.4 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: t = 0.2039 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.
Answer: i = 0.8904 A
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.
Before t = 0 both resistors are in the loop; afterwards the closed switch bypasses R₂.
Answer: φ = 62.12 °
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 = 1.988 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: cos φ = 0.9932
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 = 272 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 = 49.37 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: Cph = 725.4 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: a = 8
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.
Answer: Sη,max = 163.3 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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