Respuesta :
Answer:
The resulting time constant will be [tex]\bf{5.09~ms}[/tex].
Explanation:
Given:
the time constant of the RC circuit, [tex]\tau = 8~ms[/tex]
The value of the capacitor in the circuit, [tex]C_{1} = 4~\mu F[/tex]
The value of addition capacitor added to the circuit, [tex]C_{2} = 7~\mu F[/tex]
The value of the time constant for a series RC circuit is give by
[tex]\tau = RC[/tex]
So the value of the resistance in the circuit is
[tex]R &=& \dfrac{\tau}{C}\\&=& \dfrac{8 \times 10^{-3}~s}{4 \times 10^-6~F}\\&=& 2000~\Omega[/tex]
When the capacitor [tex]C_{2}[/tex] is added to the circuit, the net value of the capacitance in the circuit is
[tex]C &=& \dfrac{C_{1}C_{2}}{C_{1} + C_{2}}\\&=& \dfrac{4 \times 7}{4 + 7}~\mu F\\&=& \dfrac{28}{11}~\mu F[/tex]
So the new time constant will be
[tex]\tau_{n} &=& (2000~\Omega)(\dfrac{28}{11} \times 10^{-6})~s\\&=& 5.09~ms[/tex]
