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Physics
In the circuit shown in figure, the charge on the 5 μ F capacitor will be
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Q. In the circuit shown in figure, the charge on the $5\, \mu F$ capacitor will be
Electrostatic Potential and Capacitance
A
$4.5\, \mu C$
23%
B
$9 \,\mu C$
21%
C
$15\, \mu C$
38%
D
none of these
18%
Solution:
Potential difference across upper branch is $6 \,V$.
$V_{2}=V\left(\frac{C_{1}}{C_{1}+C_{2}}\right)=10\left[\frac{3}{3+7}\right]=3 \,V$
Hence, potential difference across $5 \,\mu F$ capacitor is $3 \,V$.
Thus, charge in $5 \,\mu F$ capacitor is
$Q_{5 \mu F }=5 \times 3=15\, \mu C$