JEE Main 2024 Physics question
An alternating emf E = 110√2 sin 100t volt is applied to a capacitor of 2μF, the rms value of current in the circuit is ………. mA.
Answer: 22
Solution
= 1/(100 × 2 × 10⁻⁶) = 5000 Ω; = 110/5000 = 22 mA.
An alternating emf E = 110√2 sin 100t volt is applied to a capacitor of 2μF, the rms value of current in the circuit is ………. mA.
= 1/(100 × 2 × 10⁻⁶) = 5000 Ω; = 110/5000 = 22 mA.