When the applied potential difference across a circuit element is increased linearly with time as shown in the second figure, the current in the circuit remains constant with time as shown in the first Figure. At t = 4 s the lead wire connecting the element to the source breaks. What will be the voltage across the element after t = 4 s? Can you identify the element?

Question

G1 Physics · Accepted Answer

Image For a capacitor, I = C dV/dt. A linearly increasing V gives constant I. At t = 4 s the voltage is 4 V. When disconnected, charge remains on an ideal capacitor, so voltage remains 4 V. The element is a capacitor.