In the AC circuit shown in the figure, a pure capacitor of capacitance C is connected across an alternating source. If the instantaneous current flowing in the circuit is given by I = I₀ t , the instantaneous voltage V_C across the capacitor is:

Step-by-step solution

For a pure capacitor, the relationship between the instantaneous current I and the charge q is given by: I = dq dt Since the charge on the capacitor is q = C V_C , we have: I = C dV_C dt Rearranging the equation to solve for V_C gives: dV_C = I C dt Integrating both sides with respect to time t : V_C = I C dt Given that the instantaneous current is I = I₀ t , substituting this into the integral yields: V_C = I₀ t C dt V_C = I₀ C t dt V_C = I₀ C ( t ) V_C = I₀ C t Answer: I₀ C t