Question

A capacitor with capacitance C = 7.00×10-3 F is initially uncharged. It is in series with a source of Emf of 7.00 volts, a resistor R, and a switch as shown in the figure below.

At t=0, the switch is closed. The graph below shows the potential difference across the capacitor as function of t, the time elapsed since the switch closed.

1.) Calculate the value of R. Note that the curve passes through a grid intersection point. (in ohm)

2.) What can you tell about the maximum power dissipated in R? The maximum power dissipated in R occurs at ...

a.) t=0

b.) t=infinity

c.) t= RC

Answer 1

Data Given

V = 7 V, C = 7*10^-3 F, R = ?

We know that time constant of a CR circuit is given by

$\tau = CR$ , it is the time in which 63% of the capacitor get charged

Grid intersection from where the curve passes is (6,6) So 6 s is the time constant.

So

$6 s = 7.00\times 10^{-3}F \times R$

$R= \frac{\tau }{C}= \frac{6s}{7.00\times 10^{-3}F } = 857.14 \Omega$

Part 2) Maximum power Dissipated

a) at t = 0

Current is maximum i.e. $I = \frac{V}{R}= \frac{7V}{857.14 \Omega } = 8.167 mA$

Pover Dissipated

$P = VI = 7 V \times 8.167\times 10^{-3}A = 57.169\times 10^{-3}W$

b) at t = infinity . I = 0

So Power dissipated

$P = VI = 7 V \times 0 A = 0 W$

c) at t = RC

$i = \frac{V}{R}e^{\frac{-t}{RC}}= \frac{7 V}{857.14}\times e^{\frac{-RC}{RC}}=\frac{ 8.167 mA}{e}= 3.00mA$

$P = Vi= 1 V \times 3.00\times 10^{-3}A = 3\times 10^{-3}W$