Question

a circuit includes a capacitor that charges over time. if the resistance of the circuit is 3.0 ω, the capacitance is 2.2 f, and the potential of the circuit is 5.0 v, what is the charge on the capacitor after a time of 10.0 seconds, assuming the capacitor is initially uncharged? o 11 c o 1.5 c o 2.4 c o 8.6 c

Explanation:

Step1: Recall the formula for charging a capacitor in an RC - circuit

The formula for the charge on a capacitor as a function of time in an RC - circuit is $q(t)=C\mathcal{E}(1 - e^{-\frac{t}{RC}})$, where $C$ is the capacitance, $\mathcal{E}$ is the emf of the battery, $R$ is the resistance, and $t$ is the time.

Step2: Identify the given values

We are given that $\mathcal{E}=5.0\ V$, $R = 3.0\ \Omega$, $C=2.2\ F$, and $t = 10.0\ s$.

Step3: Calculate the time - constant $\tau$

The time - constant $\tau=RC$. Substituting the values of $R$ and $C$, we get $\tau=(3.0\ \Omega)\times(2.2\ F)=6.6\ s$.

Step4: Calculate the exponent term

Calculate $e^{-\frac{t}{RC}}=e^{-\frac{10.0}{6.6}}\approx e^{- 1.515}\approx0.22$.

Step5: Calculate the charge $q$

Using the formula $q(t)=C\mathcal{E}(1 - e^{-\frac{t}{RC}})$, substitute the values: $q=(2.2\ F)\times(5.0\ V)\times(1 - 0.22)=(2.2\times5.0)\times0.78\ C = 8.58\ C\approx8.6\ C$.

Answer:

$8.6\ C$