Question

Question: An emf source with E = 120 V, a resistor with R = 80.0 Ω, and a capacitor with C = 4.00 µF are connected in series. As the capacitor charges, when the current in the resistor is 0.900 A, what is the magnitude of the charge on each plate of the capacitor?

Short answer

Answer:

When an emf source with and a resistor with and a capacitor of are connected in series and the capacitor gets charges when the current in the resistor is then the magnitude of the charge in each plate capacitor is

Step-by-step solution

EMF of the battery

Electromotive force (Emf) is the total amount of electrical potential energy a battery can supply before it is connected to an external circuit

Potential difference across the circuit

We know that the charges on the plates of the capacitor depend on the voltage across both the plates and it is given by:

$Q=C{V}_C$

Whereis the capacitance. To find the value of we need to determine. To determine the we will use the loop rule.

$$\begin{gathered} \epsilon -IR-V_C=0\backslash \mathrm{hfill} \\ {V}_C=\epsilon -IR \end{gathered}$$

Calculating the Voltage

Now in the above equation we put the values and get:

$$\begin{gathered} V_C=\epsilon -IR \\ =(120V)-(0.900A)(80\Omega ) \\ =48V \end{gathered}$$

Now, putting the values of andinto the equation we get:

$$\begin{gathered} Q=C{V}_C \\ =(4\mu F)(48V) \\ =192\mu C \end{gathered}$$

Therefore, the magnitude of the charge is$192\mu C$