MHT CET202522 Apr 2025Morning 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
- A4 ~F , 20 ~A
- B5 ~F , 10 ~A
- C2 ~F , 20 ~A
- D8 ~F , 12 ~A
Correct answer
A. 4 ~F , 20 ~A
Step-by-step solution
Maximum current in an RLC series circuit occurs at resonance when X_L = X_C , making the impedance purely resistive with Z = R and current I₀ = V₀ R . Given f = 50 Hz and ^2 = 10 , angular frequency is = 2 f = 100 rad/s. From L = 1 C , capacitance is C = 1 ^2 L = 1 (100 )^2 (2.5) . Substitute ^2 = 10 to find C = 1 1000000 2.5 = 1 250000 = 4 F . Maximum current is I₀ = V₀ R = 230 11.5 = 20 A . Thus, option A is correct.