Answer to Question #155031 in Electric Circuits for Jerick

Question

1. The circuit shown in Figure 28.5 is constructed using a 6 V battery, a 3 kΩ resistor, and a 6 kΩ resistor. Initially, both switches are open and the capacitor is uncharged.

Switch A is closed and the capacitor begins to charge.

(a) After 1 2 seconds the potential difference across the capacitor is 3.78 V. What is the capacitance of the capacitor?

After several minutes switch A is opened.

(b) What is the approximate potential difference across the capacitor?

Switch B is now closed and the capacitor begins to discharge.

(c) What is the characteristic time for discharging this capacitor?

(d) What will the current through the 6 kΩ resistor be after 2 seconds?

(e) What will the charge on the capacitor be after 3 seconds?

Answer 1

a) $q=UC$

$q=q_{max}(1-e^{-t/\tau})$

$q_{max}=EC$

$\tau=RC$

We have:

$E=6 V, U=3.78V, R=3+6=9k\Omega, t=1.2s$

Then:

$U=E(1-e^{-t/\tau})$

$-t/\tau=ln(1-U/E)$

$\tau=-t/ln(1-U/E)=-1.2/ln(1-3.78/6)=1.2s$

$C=\tau/R=1.2/9=0.13\mu F$

b) $\tau=RC=9\cdot0.13=1.2s$

c) $I=I_0e^{-t/\tau}$

$I_0$ is current at time $t=0$

$I_0=E/R=6/9=0.67mA$

$I=0.67e^{-2/1.2}=0.13mA$

d) $q=q_0e^{-t/\tau}=ECe^{-t/\tau}$

$q=6\cdot0.13e^{-3/1.2}=0.06\mu C$