An inductor of inductance 2 H is connected in series with a resistance, a variable capacitor and an a.c. source of 10 kHz. The value of capacitance for which maximum current is drawn in to the circuit is 1 x F , where the value of x is (Take ^2 = 10 )

Step-by-step solution

Maximum current is drawn in a series LCR circuit at resonance. At resonance, the inductive reactance equals the capacitive reactance, so L = 1 C . C = 1 ^2 L = 1 (2 f)^2 L Given L = 2 H = 2 10⁻⁶ H and f = 10 kHz = 10^4 Hz . C = 1 4 ^2 f^2 L Substituting the values and using ^2 = 10 : C = 1 4 10 (10^4)^2 2 10⁻⁶ C = 1 40 10^8 2 10⁻⁶ C = 1 80 10^2 = 1 8000 F Comparing with C = 1 x F , we get x = 8000 . Answer: 8000