Question

Part A Constants A capacitor is charged to a potential of 12.0 V and is then connected to a voltmeter having an internal resistance of 3.50 M2. After a time of 3.90 s the voltmeter reads 2.8 V. What is the capacitance? 0 AEO O ? You may want to review (Pages 862-866). C- For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Discharging a capacitor. Submit Request Answer Part B What is the time constant of the circuit? VO AP = Om ? Submit Request Answer

Answer 1

The voltage across the capacitor in a discharging RC circuit is given by,

$V_C\left(t \right )=V_0exp\left(-\frac{t}{RC} \right )$

where, V₀ is the maximum charge on the capacitor. We can rearrange the above equation as,

$\frac{V_C\left(t \right )}{V_0}=exp\left(-\frac{t}{RC} \right )$

Taking natural logarithm on both sides we get,

$-\frac{t}{RC}=\ln\frac{V_C\left(t \right )}{V_0}$

or, $C=-\frac{t}{R\ln\frac{V_C\left(t \right )}{V_0}}$

In the problem, V₀=12 V, R= 3.5 MOhm. At t=3.9s, V_C(t)=2.8V

Therefore, $C=-\frac{3.9}{3.5\times10^6 \times \ln\frac{2.8}{12}}\approx 766\;nF$

Time constant $\tau=RC=3.5\times 10^6 \times 766 \times 10^{-9}\;sec=2.681sec$