Question

the initial charge on a capacitor is 5.0 c, the resistance is 4.5 ω, and the capacitance is 0.10 f. what is the charge on the capacitor after 0.50 seconds of discharging? o 5.0 c o 1.6 c o 0.16 c o 4.0 c

Explanation:

Step1: Recall RC - discharge formula

The formula for the charge on a capacitor during discharge is $q = q_0e^{-\frac{t}{RC}}$, where $q_0$ is the initial charge, $R$ is the resistance, $C$ is the capacitance, $t$ is the time, and $q$ is the charge at time $t$.

Step2: Calculate the time - constant $\tau$

The time - constant $\tau=RC$. Given $R = 4.5\Omega$ and $C = 0.10F$, then $\tau=4.5\times0.10 = 0.45s$.

Step3: Substitute values into the formula

We have $q_0 = 5.0C$, $t = 0.50s$, and $\tau=0.45s$. Substitute into $q = q_0e^{-\frac{t}{RC}}=q_0e^{-\frac{t}{\tau}}$. So $q = 5.0\times e^{-\frac{0.50}{0.45}}$.
Calculate $e^{-\frac{0.50}{0.45}}=e^{- 1.11}\approx0.329$. Then $q = 5.0\times0.329 = 1.645C\approx1.6C$.

Answer:

$1.6C$