Question

Discharging RC circuit. • In the RC circuit shown, the battery has fully charged the capacitor, so Q0 = C E. Then at t = 0 the switch is thrown from position a to b. The battery emf is 20.0 V, and the capacitance C = 1.02 μF. The current I is observed to decrease to 0.50 of its initial value in 40 μs. (a) What is the value of Q, the charge on the capacitor, at t = 0? (b) What is the value of R?  (c) What is Q at t = 60 μs? 

Answer 1

a) At $t=0$ the value of the charge on the capacitor will be:

$$Q=CE=1.02\cdot10^{-6}\ F\cdot20\ V=20.4\cdot10^{-6}\ C=20.4\ \mu C.$$

b) Let's write the equation for current:

$$I(t)=I_0e^{-\dfrac{t}{RC}}.$$

Since, $I=0.5I_0$, we get:

$$0.5I_0=I_0e^{-\dfrac{t}{RC}},$$ $$ln(0.5)=ln(e^{-\dfrac{t}{RC}}),$$ $$-\dfrac{t}{RC}=ln(0.5)=-ln2,$$ $$R=\dfrac{t}{Cln2}=\dfrac{40\cdot10^{-6}\ s}{1.02\cdot10^{-6}\ F\cdot ln2}=57\ \Omega.$$

c) Let's first find the time constant:

$$\tau=RC=57\ \Omega\cdot1.02\cdot10^{-6}=58\cdot10{-6}\ s=58\ \mu s.$$

Then, we can find $Q$ at $t=60\ \mu s$:

$$Q=Q_0e^{-\dfrac{t}{\tau}}=20.4\cdot10^{-6}\ C\cdot e^{-\dfrac{60\ \mu s}{58\ \mu s}}=7.2\ \mu C.$$