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3. current electricity — Physics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 SciencePhysics3. current electricity1 Mark
MCQ
Circuit shown is in steady state, now when switch is closed, galvanometer shows no deflection, then correct relation is
  • ✓
    $\frac{{{R_1}}}{{{R_2}}} = \frac{{{C_2}}}{{{C_1}}}$
  • B
    $\frac{{{R_1}}}{{{R_2}}} = \frac{{{C_1}}}{{{C_2}}}$
  • C
    $R_1R_2 = C_1C_2$
  • D
    ${R_1}\sqrt {{C_1}}  = {R_2}\sqrt {{C_2}} $
✓

Answer

Correct option: A.
$\frac{{{R_1}}}{{{R_2}}} = \frac{{{C_2}}}{{{C_1}}}$
a
$\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ are in series with battery so charge on both will be same

let it is $=\mathrm{q}$

and let current in resistance branch $=\mathrm{i}$

$\therefore \mathrm{V}_{\mathrm{C}_{1}}=\mathrm{i} \mathrm{R}_{1} \Rightarrow \frac{\mathrm{q}}{\mathrm{C}_{1}}=\mathrm{i} \mathrm{R}_{1} \Rightarrow \mathrm{q}=\mathrm{ir}_{1} \mathrm{C}_{1}$

similarly $\mathrm{V}_{\mathrm{C}_{2}}=\mathrm{i} \mathrm{R}_{2} \Rightarrow \frac{\mathrm{q}}{\mathrm{C}_{2}}=\mathrm{i} \mathrm{R}_{2} \Rightarrow \mathrm{q}=\mathrm{i} \mathrm{R}_{2} \mathrm{C}_{2}$

$\therefore \quad \mathrm{iR}_{1} \mathrm{C}_{1}=\mathrm{i} \mathrm{R}_{2} \mathrm{C}_{2}$

$\Rightarrow \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}}$

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