Alternating Current — Questions & Answers

Sample questions from Alternating Current

1. Q1. In a series LCR circuit, removing the capacitor makes the current lag the voltage by 60°, while removing the inductor instead makes the current lead by 30°. With all three elements present, the circuit is:

   • A.At resonance (power factor 1)
   • B.Inductive, with tan φ = 2/√3✓
   • C.Capacitive, with tan φ = 2/√3
   • D.Inductive, with tan φ = √3
   Solution

   Without C: tan 60° = XL/R gives XL = √3 R. Without L: tan 30° = XC/R gives XC = R/√3. With both: XL − XC = R(√3 − 1/√3) = R(3 − 1)/√3 = 2R/√3 > 0, so the circuit is inductive with tan φ = 2/√3 (φ ≈ 49°).

2. Q2. A capacitor and an inductor are connected in PARALLEL (rather than series) across an AC source, near their own parallel-resonance condition. At parallel resonance, the combination's overall impedance to the source is:

   • A.Always exactly zero
   • B.At its minimum value, same as series resonance
   • C.At its MAXIMUM value (opposite to the series-resonance case, where impedance is minimum)✓
   • D.Completely independent of the frequency
   Solution

   For a parallel LC combination (an 'anti-resonant' or 'rejector' circuit), at the resonant frequency the combination's overall impedance to the external source is at its MAXIMUM (opposite in behaviour to the SERIES LCR case, where impedance is minimum at resonance) -- this means the combination draws MINIMUM current from the source at parallel resonance, a useful property exploited in certain filter and tuning applications (a 'rejector' circuit that blocks a specific frequency, rather than a series circuit's 'acceptor' behaviour that favours a specific frequency). A student who assumes parallel and series resonance behave identically (both minimum impedance) picks B.

3. Q3. A transformer steps 3300 V down to 220 V. Its primary draws 5 A and its efficiency is 96%. The secondary current is:

   • A.69.1 A
   • B.72 A✓
   • C.75 A
   • D.15 A
   Solution

   Primary power = 3300 x 5 = 16500 W. Output power = 0.96 x 16500 = 15840 W, so Is = 15840/220 = 72 A. The ideal (100% efficient) answer would be 75 A; efficiency reduces it.

4. Q4. A choke coil designed for a 220 V, 50 Hz AC supply is mistakenly connected to a 220 V DC source. The most likely result is:

   • A.Nothing happens, because a coil blocks DC
   • B.A very large current flows and the coil overheats, since only its small resistance limits DC✓
   • C.The coil works normally, since 220 V is 220 V
   • D.The current is limited by the inductive reactance as before
   Solution

   A choke limits AC current by its large reactance ωL while having a small ohmic resistance. For DC, ω = 0 and XL = 0, so after the brief transient the current is V/R with R very small: a dangerously large current flows and the coil may burn out. A choke controls AC current with little power loss but cannot control DC.

5. Q5. A pure inductor is connected to an AC source of fixed RMS voltage. If the source frequency is doubled, the RMS current becomes:

   • A.Double
   • B.Half✓
   • C.Four times
   • D.Unchanged
   Solution

   Irms = Vrms/XL = Vrms/(2πfL). Doubling f doubles XL, so at fixed Vrms the current halves. An inductor increasingly opposes higher-frequency currents.

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