Question

A series RC circuit, which is made from a battery, a switch, a resistor, and a 4-μF capacitor, has a time constant of 8 ms. If an additional 7-μF is added in series to the 4-μF capacitor, what is the resulting time constant?

Answer 1

Answer:

The resulting time constant will be $\bf{5.09~ms}$.

Explanation:

Given:

the time constant of the RC circuit, $\tau = 8~ms$

The value of the capacitor in the circuit, $C_{1} = 4~\mu F$

The value of addition capacitor added to the circuit, $C_{2} = 7~\mu F$

The value of the time constant for a series RC circuit is give by

$\tau = RC$

So the value of the resistance in the circuit is

$R &=& \dfrac{\tau}{C}\\&=& \dfrac{8 \times 10^{-3}~s}{4 \times 10^-6~F}\\&=& 2000~\Omega$

When the capacitor $C_{2}$ is added to the circuit, the net value of the capacitance in the circuit is

$C &=& \dfrac{C_{1}C_{2}}{C_{1} + C_{2}}\\&=& \dfrac{4 \times 7}{4 + 7}~\mu F\\&=& \dfrac{28}{11}~\mu F$

So the new time constant will be

$\tau_{n} &=& (2000~\Omega)(\dfrac{28}{11} \times 10^{-6})~s\\&=& 5.09~ms$