MCQ
In given circuit
- ACharge on $C_1$ is zero
- ✓Charge on $C_1$ is $80\, \mu C$
- CCharge on $C_2$ is $40\, \mu C$
- DCharge on $C_2$ is $20\, \mu C$
✓
Answer
Correct option: B.
Charge on $C_1$ is $80\, \mu C$
b
$\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{a}}=\frac{120}{6} \times 1=20 \mathrm{\,V}$
$\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{a}}=\frac{120}{6} \times 1=20 \mathrm{\,V}$
$\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{b}}=\frac{120}{6} \times 2=40 \mathrm{\,V}$
$\therefore \mathrm{V}_{\mathrm{a}}-\mathrm{V}_{\mathrm{b}}=40-20=20 \mathrm{\,V}$
charge on $\mathrm{C}_{1}$ is $\mathrm{q}_{1}=\mathrm{C}_{1}\left(\mathrm{V}_{\mathrm{a}}-\mathrm{V}_{\mathrm{b}}\right)$
$=40 \times 10^{-6} \times 20=80 \,\mu C$
$\mathrm{V}_{\mathrm{c}}=\mathrm{V}_{\mathrm{d}}$ and $\mathrm{V}_{\mathrm{e}}=\mathrm{V}_{\mathrm{f}}$
so charge on $\mathrm{C}_{2}$ and $\mathrm{C}_{3}$ are zero.
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