Thread: Voltage drop

A 6V DC voltage source is connected across both a 5 Kohm resistor and a 10nF capacitor in series. The voltage drop across the resistor is equal to what

The expression for the voltage across the capacitor when charging, is almost the same as when it is discharging (as in your other question), except the time will not have the negative co-efficient. This is due to the application of Kirchoff's second law in the opposite direction.

The expression is therefore:

V_c = V_0 *e^(t/RC)

(this expression can be used alongside the expression Q = Q_0 *e^(t/RC) since V = Q/C )

For a DC supply, the sum of the voltage drops of ALL components in the circuit (including internal resistance of the battery) is equal to the EMF of the supply.

That is, in this case:

V_R = V_supply - V_c