Question

The circuit in the drawing shows two resistors, a capacitor, and a battery. When the capacitor is fully charged, what is the magnitude q of the charge on one of its plates? (Assume that R1 = 9.0 Ω, R2 = 2.5 Ω, and C = 8.0 µF.)

Answer 1

The question is incomplete, here is the complete question "The circuit in the drawing shows two resistors, a capacitor, and a battery. When the capacitor is fully charged, what is the magnitude q of the charge on one of its plates? (Assume that V=12v, R1 = 9.0 Ω, R2 = 2.5 Ω, and C = 8.0 µF.)

"

Answer =10.4µC

Step-by-step explanation:

See the attached file for the circuit diagram.

To determine the charge on the capacitor's plate, we use the formula Q=CV were

Q=quantity of charge

C=capacitance of the capacitor and

v= voltage across the capacitor.

Note: when Resistors are arrange in series, different voltage value pass through them depending on there values and when arranged in parallel irrespective of there values the voltage across them is the same.

Going by this theory, we define the voltage value across the resistor(R2) in parallel to the capacitor,since the same value of voltage will be found across them.

from the second attached file we write the voltage divider rule we have

`V_(2)=(VR_(2) )/(R_(1)+R_(2)) \\V_(2)=(12*2.5)/(9+2.5) \\V_(2)=2.6V`.

Hence we can conclude that the voltage in the capacitor also is 2.6v since it is in parallel to the second resistor.

Now using the earlier stated formula i.e Q=CV

Where C=8µf

`Q=8*10^(-6)* 2.6\\Q=20.8*10^(-6)C\\`

Hence the charge across one of its plate is Q/2

`Q=10.4*10^(-6)C\\`

or 10.4µC