Question

Determine the charge on the capacitor in the following circuit:
(A) 200 μC (B) 60 μC
(C) 10 μC (D) 2 μC

Answer 1

The charge on the capacitance can be calculated as below:

• First calculate the total resistance in the circuit that is 6+2=8 Ω
• Then calculate the current with voltage 72V that is I=V/R
• I=72/8 that is I=9
• Now voltage at 10Ω= 20V
• The formula of charge on capacitance is q=CV
• Putting the values q=10×20=200μC

Answer 2

The charge on the capacitor is 200 μ C

Step by step explanation:

Observing the given circuit, to get the charge on the capacitor

The formula for charge = $\textbf{\Large q = CV}$                             (equation 1)

From figure, we know that C = 10 μF                               (given)

Here V is the potential difference between capacitor

As we know that potential difference in branches of parallel wires is same.

So the potential difference in 10 ohm resistance is same as that of the capacitor.

To find the pot. diff. on capacitor , we need to first get the current through 10 ohm resistance and to find the current through it, we need to find the equivalent resistance of circuit.

For equivalent resistance  10 and 2 ohm resistors are in series so total resistance by them = 10 + 2 = 12 ohm.

Then 4 ohm and this 12 ohm are in parallel so total resistance = $\frac{4\times 12}{4+12}$  = 3 ohm

This 3 ohm and 6 ohm resistance are in series so

Equivalent resistance =  6 + 3 = 9 ohm

So Total current = $I = \frac{V}{R_e_q} = \frac{72}{9} = 8\ ampere$

So 8 ampere current runs through 6 ohm resistance and potential drop through 6 ohm resistance=

$V = I \times R$ = $8 \times 6 = 48$ Volts

Remaining voltage = 72 - 48 = 24 volts is the voltage through both parallel wires.

Now we can see that 24 volts is to drop in 2 and 10 ohm resistance.

So 4 volts will drop in 2 ohm resistance and

20 v will drop in 10 ohm resistance as well as through capacitor (because both are parallel)

$\textbf{Total voltage through capacitor V = 20 Volts}\\\\\textbf{Total capacitance of capacitor C = 10}\mu C\\\\\textbf{\Large So Total charge in capacitor q = CV =10}\times \textbf{\Large 20 = 200}\mu C$

Correct option is (A)