Question

A circuit contains a resistor in series with a capacitor, the series combination being connected across the terminals of a battery. The time constant for charging the capacitor is 4.6 s when the resistor has a resistance of 1.4 x 104 Ω. What would the time constant be if the resistance had a value of 5.6 x 104 Ω?

Answer 1

Answer:

18.4 s

Explanation:

The time constant of an RC circuit is given by

$\tau = RC$

where

R is the resistance

C is the capacitance

For the first circuit we have

$\tau = 4.6 s$

$R=1.4\cdot 10^4 \Omega$

So we can find the capacitance

$C=\frac{\tau}{R}=\frac{4.6 s}{1.4\cdot 10^4 \Omega}=3.29\cdot 10^{-4} F$

Now in the second circuit, the new resistance is

$R=5.6\cdot 10^4 \Omega$

So the new time constant will be

$\tau = RC=(5.6\cdot 10^4 \Omega )(3.29\cdot 10^{-4} F)=18.4 s$