The maximum flow of current through the resistor will occur at the instant the switch is closed, and will be:
V | = | 10 V | = 1 mA |
R | 10 kΩ |
The maximum charge on the capacitor will be achieved when the potential difference across the capacitor is equal to the terminal voltage of the battery, and will be:
Q = V × C = 10 V × 10 μF = 100 μC
The time constant for the circuit is 100 μF × 50k Ω = 5 (seconds), therefore the capacitor voltage VC is given by:
VC = 0.63 × 12 V = 7.56 V
and since VSUPPLY = VC + VR:
VR = VSUPPLY - VC = 12 V - 7.56 V = 4.44 V
When the switch is moved to position B, the battery is no longer part of the circuit and the capacitor starts to discharge through the resistor. Five seconds after the switch is moved to position B, the capacitor voltage will have fallen to 0.37 of the voltage it had at the instant the switch was moved from position A to position B (7.56 V). The new instantaneous voltage is therefore given by:
VC = 0.37 × 7.56 V = 2.7972 V
The instantaneous voltage drop across the resistor is the same as the capacitor voltage, and is given by:
VR = VC = 2.7972 V
C | R | Time Constant |
---|---|---|
1 μF | 1 MΩ | 1 second |
1000 μF | 1 kΩ | 1 second |
0.22 μF | 150 kΩ | 0.033 seconds |
10 μF | 47 kΩ | 0.47 seconds |
The maximum current flows at the instant the circuit is first connected to the battery, and is given by:
I = | VSUPPLY | = | 1.5 V | = 0.01 mA |
R | 150 kΩ |
The maximum charge on the capacitor occurs when the potential difference across the capacitor is equal to the supply voltage, and is given by:
Q = VSUPPLY × C = 1.5 V × 1000 μF = 1500 μC
One volt is exactly two thirds of the supply voltage (1.5 V). Since the capacitor voltage will have reached approximately two thirds of the supply voltage after one time constant (T), the approximate time required for the capacitor to reach a potential of 1.0 V will be:
T = C × R = 1000 μF × 150 Ω = 0.15 seconds