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qwelly [4]
3 years ago
13

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 a

n additional 7-μF is added in series to the 4-μF capacitor, what is the resulting time constant?
Physics
1 answer:
Verizon [17]3 years ago
7 0

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

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