A source of alternating emf = ₀ ( t) is connected to a capacitor. Then the instantaneous current in the circuit is:

Step-by-step solution

The alternating emf is given by = ₀ ( t) . The charge on the capacitor at any instant is q = C = C ₀ ( t) . The instantaneous current I in the circuit is the rate of flow of charge, given by I = dq dt . Differentiating q with respect to time t , we get: I = d dt (C ₀ ( t)) = C ₀ ( t) Using the trigonometric identity ( t) = ( t + 2 ) , the current can be written as: I = I₀ ( t + 2 ) where I₀ = ₀ C is the peak current. This shows that in a purely capacitive circuit, the current leads the voltage by a phase angle of 2