Question

A series RC circuit, which is made from a battery, a switch, a resistor, and a 3.0 µF capacitor, has a time constant of 10.0 ms. If an additional 8.0 µF is added in series to the 3.0 µF capacitor, what is the resulting time constant?

Answer 1

Answer:

New time constant is 7.27 ms

Explanation:

Time constant of RC circuit is given as

$\tau = RC$

now we know that time constant is given as 10 ms for capacitor of 3 micro farad

then it is given as

$10\times 10^{-3} s = (3\mu F)(R)$

$R = 3333.33 ohm$

now the capacitor is connected with 8 micro farad capacitor in series then we have

$C = \frac{(3\mu F)(8\mu F)}{3\mu F + 8\mu F}$

$C = 2.18 \mu F$

now the new time constant is given as

$\tau' = (2.18\mu F)(3333.3)$

$\tau' = 7.27 ms$