A capacitor stores energy of 1.25 J.If when it is connected by means of a switch to a 15-$\varOmega$ resistor it discharges to half this value in 15 s,how much charge flowed off the capacitor in this time?
A)0.34 C
B)0.79 C
C)0.95 C
D)1.9 C

Question

Quizplus · Accepted Answer

The energy stored in a capacitor can be calculated as $E=\frac{1}{2}CV^2$, where $C$ is the capacitance and $V$ is the voltage across the capacitor.
We are given that the capacitor stores 1.25 J, so we can solve for the initial voltage: $1.25=\frac{1}{2}(C)(V_{0}^2)$. Since $C$ is not given, we cannot solve for $V_{0}$.
Next, we are told that the capacitor discharges to half its initial value, so the final voltage is $V_{f}=\frac{V_{0}}{2}$.
We are also given the resistance and time, so we can use the equation for the discharge of a capacitor through a resistor: $V_{f}=V_{0}e^{-\frac{t}{RC}}$, where $R$ is the resistance and $C$ is the capacitance. Since we don't know $C$, we can rearrange the equation to solve for $RC$: $RC=-\frac{t}{\ln{\frac{V_{f}}{V_{0}}}}=-15\ln{2}=10.4\Omega	imes F$
Now we can solve for $C$ using $RC=\frac{1}{C}$: $C=\frac{1}{RC}=0.096 F$.
Finally, we can use the equation $Q=CV$ to solve for the charge on the capacitor when it is half discharged: $Q=\frac{1}{2}(C)V_{f}=0.048 C$. Therefore, the amount of charge that flowed off the capacitor in 15 s is $Q_{0}-Q=1.25-0.048=1.202 C$. However, the question asks for the amount of charge that flowed off the capacitor in 15 s, so we divide by 2 to get the answer of 0.79 C.

A Capacitor Stores Energy of 1 Ω\varOmegaΩ Resistor It Discharges to Half This Value in 15 S,how

A Capacitor Stores Energy of 1 $\varOmega$ Resistor It Discharges to Half This Value in 15 S,how