A 110 V, 50 Hz, AC source is connected in the circuit (as shown in figure). The current through the resistance 55 Ω, at resonance in the circuit, will be __________ A.

Solution

Physics sir · 4 steps

L and C sit in parallel here, so at resonance the tank blocks the current instead of letting it through.

1. Spot the connection. The inductor and the capacitor are in parallel with each other. That parallel pair is then in series with the 55 Ω resistor and the source.
2. Branch currents cancel. Both branches have the same voltage across them. The current in L lags this voltage by 90°, and the current in C leads it by 90°, so the two currents are exactly opposite in phase.

   Iₙₑₜ = V(ω C - 1/(ω L)) = 0 at ω = 1/√LC

3. Infinite impedance. At resonance X_L = X_C, so the two branch currents are equal in size and cancel. The ideal LC tank then has infinite impedance and draws no current from the rest of the circuit.
4. So. No current flows through the series 55 Ω resistor, so the answer is 0 A.

The common slip: 2. Treating this as a series LCR circuit gives I = 110/55 = 2 A. That is wrong because L and C are in parallel, and a parallel LC tank blocks current at resonance.

The idea to keep. Series LCR at resonance has minimum impedance and maximum current. A parallel LC at resonance has maximum (ideally infinite) impedance, so the line current drops to zero. Check the wiring first.

Answer from the NTA final answer key. Solution by PrepWiser.