In the shown AC source, the voltage is given as V = 20 cos 2000t. Neglecting source resistance, the voltmeter and ammeter readings will be:
0 V, 1.4 A
We analyze the given circuit using impedance calculations.
- Given AC source: V = 20 cos(2000t).
- Inductive reactance: XL = ωL = (2000) × (5 × 10-3) = 10 Ω.
- Capacitive reactance: XC = 1 / (ωC) = 1 / (2000 × 50 × 10-6) = 10 Ω.
- Net reactance: Xnet = XL - XC = 10 - 10 = 0 Ω.
- Total resistance in series: R = 6Ω + 4Ω = 10Ω.
- Total impedance: Z = √(R² + Xnet²) = √(10² + 0²) = 10 Ω.
- Current: I = V / Z = 20 / 10 = 2 A (RMS value = 1.4 A).
- Voltage across inductor and capacitor (voltmeter reading) = 0 V due to resonance.
Thus, the correct answer is (a).
Important Questions from Alternating Current
The same current is flowing in two AC circuits. The first circuit contains a pure inductor and the second, a capacitor. If the frequency of the AC is increased, then the current will:
A 25 μF capacitor, a 0.10 H inductor, and a 25 Ω resistor is connected in series with an AC source of emf ε = 310 sin 314t. What is the frequency of the AC source?
The same current is flowing in two AC circuits. The first circuit contains a pure inductor and the second, a capacitor. If the frequency of the AC is increased, then the current will:
A 25 μF capacitor, a 0.10 H inductor, and a 25 Ω resistor is connected in series with an AC source of emf ε = 310 sin 314t. What is the frequency of the AC source?
Match List-I with List-II:
List-I List-II (A) Impedance of a series RLC circuit at resonance (I) Voltage across L & C are 180° out of phase (B) For a series LC circuit (II) Current in L & C are 180° out of phase (C) For a parallel LC circuit (III) Minimum (D) Reactance of a capacitor in DC circuit (IV) Infinite Choose the correct answer from the options given below:
