MHT CET202520 Apr 2025Evening ShiftPhysicsAlternating CurrentActual
An a.c. e.m.f. of peak value =230 ~V and frequency 50 Hz is connected to a circuit with R =11.5 , ~L =2.5 H and a capacitor all in series. The value of capacitance is ' C ' for the current in the circuit to be maximum. The value of C and maximum current are respectively ( ^2=10 )
Options
- A2 F , 10 ~A
- B4 F , 20 ~A
- C6 F , 10 ~A
- D8 F , 20 ~A
Correct answer
B. 4 F , 20 ~A
Step-by-step solution
Resonance condition: Maximum current occurs when X_L = X_C , where L = 1 C . Angular frequency = 2 f = 2 (50) = 100 rad/s . Solving for capacitance: C = 1 ^2 L = 1 (100 )^2(2.5) . Using ^2 = 10 : C = 1 10000 10 2.5 = 1 250000 = 4 10⁻⁶ F = 4 F . At resonance, impedance equals resistance Z = R = 11.5 , so peak current I₀ = V₀ R = 230 11.5 = 20 A . These values correspond to option B: B .