b $\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{a}}=\frac{120}{6} \times 1=20 \mathrm{\,V}$ $\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{b}}=\frac{120}{6} \times 2=40 \mathrm{\,V}$ $\therefore \mathrm{V}_{\mathrm{a}}-\mathrm{V}_{\mathrm{b}}=40-20=20 \mathrm{\,V}$ 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.

Charge on $C_1$ is $80\, \mu C$

Charge on $C_2$ is $40\, \mu C$

Charge on $C_2$ is $20\, \mu C$

b $\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{a}}=\frac{120}{6} \times 1=20 \mathrm{\,V}$ $\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{b}}=\frac{120}{6} \times 2=40 \mathrm{\,V}$ $\therefore \mathrm{V}_{\mathrm{a}}-\mathrm{V}_{\mathrm{b}}=40-20=20 \mathrm{\,V}$ 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.

In given circuit

SECTION - A [PHYSICS MCQ]

GSEB - English Medium

In given circuit

ACharge on $C_1$ is zero
BCharge on $C_1$ is $80\, \mu C$
CCharge on $C_2$ is $40\, \mu C$
DCharge on $C_2$ is $20\, \mu C$

Diffcult

STD 11 Science
JEE
STD 12 - 3. current electricity
Physics

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