Question

Determine the charge on the capacitor in the following circuit

• A. $110\mu C$
  
• B. $60\mu C$
  
• C. $200\mu C$
  
• D. $2\mu C$
  

Answer 1

The correct option is C $200\mu C$


In steady state capacitor behaves as open circuit
The circuit can be reduced as following:

Hence, $R_{eq}=9\Omega$
Given $E=72V$
The current supplied by the battery

$i=\frac{72V}{9\Omega }=8A$. (ohm's law )

Thus, the current supplied by the battery is $8A$.

assume zero potential at lower potential terminal of cell
Now distributing currents using $KCL$ and writing potentials at different nodes using ohms law .

ohm's law for $6\Omega :$
$\frac{V_A-V_B}{R}=8A$

$\Rightarrow \frac{72V-V_B}{6\Omega }=8A$

$\Rightarrow V_B=24V$

ohm's law for $2\Omega :$
$\frac{V_B-V_C}{R}=2A$

$\Rightarrow \frac{24V-V_C}{2\Omega }=2A$

$\Rightarrow V_C=20V$

We can observe the potential difference across the capacitor is $20V$.

Now, voltage across capacitor ,
$V=20V$,

$C=10\mu F$

$\therefore$ charge on capacitor $q=CV$

$=10\times {10}^{-6}F\times 20V=200\mu C$

Hence, option C is correct.