JEE MAIN - Physics (2023 - 8th April Morning Shift - No. 18)
In this figure the resistance of the coil of galvanometer G is $$2 ~\Omega$$. The emf of the cell is $$4 \mathrm{~V}$$. The ratio of potential difference across $$\mathrm{C}_{1}$$ and $$\mathrm{C}_{2}$$ is:
_8th_April_Morning_Shift_en_18_1.png)
$$\frac{5}{4}$$
$$\frac{3}{4}$$
1
$$\frac{4}{5}$$
Explanation
At a steady state, no current would be flowing in capacitor circuit.
Required $=6+2+8=16 \Omega$
From Ohm's law
$$ i=\frac{\mathrm{V}}{\mathrm{R}}=\frac{4}{16}=\frac{1}{4} \mathrm{~A} $$
Voltage across AC
$$ \mathrm{V}_{\mathrm{AC}}=i(6+2)=\frac{1}{4} \times 8=2 \mathrm{~V} $$
Voltage across BD
$$ \mathrm{V}_{\mathrm{BD}}=i(2+8)=\frac{1}{4} \times 10=2.5 \mathrm{~V} $$
Now $\frac{\mathrm{V}_{\mathrm{AC}}}{\mathrm{V}_{\mathrm{BD}}}=\frac{2}{2.5}=\frac{4}{5}$
Required $=6+2+8=16 \Omega$
From Ohm's law
$$ i=\frac{\mathrm{V}}{\mathrm{R}}=\frac{4}{16}=\frac{1}{4} \mathrm{~A} $$
Voltage across AC
$$ \mathrm{V}_{\mathrm{AC}}=i(6+2)=\frac{1}{4} \times 8=2 \mathrm{~V} $$
Voltage across BD
$$ \mathrm{V}_{\mathrm{BD}}=i(2+8)=\frac{1}{4} \times 10=2.5 \mathrm{~V} $$
Now $\frac{\mathrm{V}_{\mathrm{AC}}}{\mathrm{V}_{\mathrm{BD}}}=\frac{2}{2.5}=\frac{4}{5}$
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