Question

In the circuit, R = 21 kΩ and C = 0.56 μF. The capacitor is allowed to charge fully and then the switch is changed from position a to position b. What will the voltage across the resistor be 1.7 ms later?

Answer 1

Answer:

78 V

Explanation:

The figure of the circuit is missing: find it in attachment.

At the beginning, the switch is in the position "a", so the current flows from the battery through the resistor and the capacitor. So, the capacitor charges.

The final charge on the capacitor is given by:

$Q_0 = CV_0$

where

$C=0.56 \mu F = 0.56\cdot 10^{-6}F$ is the capacitance

$V_0=90 V$ is the final potential difference across the the capacitor (equal to that of the battery)

So,

$Q_0=(0.56\cdot 10^{-6})(90)=5\cdot 10^{-5} C$

Later, the switch is moved to position b, so now the capacitor will discharge through the resistor. As a result, the current in the circuit follows the law

$I=I_0e^{-\frac{t}{RC}}$

where

$R=21 k\Omega = 21,000 \Omega$ is the resistance

And according to Ohm's law, the voltage across the resistor is given by

$V_R=IR$

Therefore,

$V_R(t)=I_0 R e^{-\frac{t}{RC}}$

However, $I_0 R=V_0$ (potential difference across the battery), so

$V_R(t)=V_0 e^{-\frac{t}{RC}}$

Therefore, substituting

t = 1.7 ms = 0.0017 s

We find:

$V_R(t)=(90)e^{-\frac{0.0017}{(21000)(0.56\cdot 10^{-6})}}=78 V$