Question 160: Electrostats & Capacitance

Question

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:

Accepted Answer

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
$$
\mathrm{V}_{\mathrm{AC}}=i(6+2)=\frac{1}{4} 	imes 8=2 \mathrm{~V}
$$
Voltage across BD
$$
\mathrm{V}_{\mathrm{BD}}=i(2+8)=\frac{1}{4} 	imes 10=2.5 \mathrm{~V}
$$
Now $\frac{\mathrm{V}_{\mathrm{AC}}}{\mathrm{V}_{\mathrm{BD}}}=\frac{2}{2.5}=\frac{4}{5}$

Answer

$$\frac{5}{4}$$

Answer

$$\frac{3}{4}$$

Answer

1

Answer

$$\frac{4}{5}$$

Electrostats & Capacitance

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:

Choose the correct answer: