Capacitors

When the capacitor in a circuit is fully charged what is the value of the current in the circuit?

Zero

When a capacitor in a circuit is fully charged what is the value of the potential difference across the plates?

The same as the EMF of the circuit.

Relationship of the charge of the capacitor to the potential difference of the capacitor.

Proportional.

What circuit is used to measure the discharge of a capacitor?

RC circuit with a switch that can either charge the capacitor or discharge the capacitor.

What equation is used to measure the discharge of a capacitor?

$$X=X_0e^{-\frac{t}{CR}}$$

What circuit is used to measure the charging of a capacitor?

An RC circuit with the resistor and capacitor in series and a switch to either turn on the circuit or turn off the circuit.

What equation is used to measure the charging of a capacitor?

X=X0(1-e^(-t/CR))

Equation for time constant

Time constant = CR

Definition of time constant

Time constant is equal to the time taken for the p.d, current or charge to decrease to 1/e of the original value.

What equation shows what factors affect capacitance

$$C=\frac{\varepsilon_0A}{d}$$

What factors affect capacitance and what is the relationship of capacitance with these factors?

Capacitance is proportional to area of the plate, A

Capacitance is inversely proportional to the separation of the plates, d

What are some uses of capacitors?

Smoothing a AC circuit to DC circuit by discharging as voltage is negative.

Photography by discharging rapidly.

Sum of capacitance for capacitors in series

1/Ct = 1/C1 + 1/C2 + …

Sum of capacitance for capacitors is parallel

Ct = C1 + C2 + …

If a capacitor is connected to a circuit and you alter either the plate area or plate separation what will happen?

C = Q/V and C = $\frac{\varepsilon_0A}{d}$

As the capacitor is connected to a circuit the voltage will not be changed.

Instead the charge of the capacitor will change.

If a capacitor is disconnected from a circuit and you alter either the plate area or plate separation what will happen?

C = Q/V and C= $\frac{\varepsilon_0A}{d}$

As the capacitor is disconnected from a circuit charge has nowhere to flow so will not be changed

Instead the voltage will change