A series LCR circuit is connected to a 45 sin ( ω t ) Volt source. The resonant angular frequency of the circuit is 10 5 rad s − 1 and current amplitude at resonance is I 0 . When the angular frequency of the source is ω = 8 × 10 4 rad s − 1 , the current amplitude in the circuit is 0 . 05 I 0 . If L = 50 mH , match each entry in List- I with an appropriate value from List- II and choose the correct option. List- I L

Step-by-step solution

Resonant angular frequency is given by, 1 L C = 10 5 1 50 × 10 - 3 C = 10 5 ⇒ C = 2 × 10 - 9 F Given: V = 45 sin ω t . Therefore, V 0 = 45 . Now, I 0 = V 0 R = 45 R . . . ii Inductive reactance, X L = ω L = 8 × 10 4 × 50 × 10 - 3 = 4000 Ω . and capacitive reactance, X C = 1 ω C = 1 8 × 10 4 × 2 × 10 - 9 = 6250 Ω . For new current amplitude, we can write 0 . 05 I 0 = 45 R 2 + X L - X C 2 ⇒ 0 . 05 I 0 = 45 R 2 + 6250 - 4000 2 ⇒ 0 . 05 × 45 R = 45 R 2 + 6250 - 4000 2 ⇒ R 2 + 6250 - 4000 2 = R 2 0 . 05 2 ⇒ R 2 + 2250 2