Question 562: Electrostats & Capacitance

Question

Two capacitors $$\mathrm{C}_1=3 \mu \mathrm{F}$$ and $$\mathrm{C}_2=2 \mu \mathrm{F}$$ are connected in series across d.c. source of $$100 \mathrm{~V}$$. The ratio of the potential across $$C_2$$ to $$C_1$$ is

Accepted Answer

$$\begin{aligned}
& \mathrm{C}_1=3 \mu \mathrm{F} 	ext { and } \mathrm{C}_2=2 \mu \mathrm{F} \
	herefore \quad & \mathrm{C}_{	ext {series }}=\frac{\mathrm{C}_1 \mathrm{C}_2}{\mathrm{C}_1+\mathrm{C}_2}=\frac{6}{5} \mu \mathrm{F} \
& 	ext { Also, } \mathrm{Q}=\mathrm{CV} \
\mathrm{Q} & =\mathrm{C}_{	ext {series }} 	imes \mathrm{V} \
& =\frac{6}{5} 	imes 100=120 \mu \mathrm{C}
\end{aligned}$$
Q will be the same across both the capacitors as they are in series.
$$	herefore \quad$$ Potential across capacitors,
$$\begin{aligned}
& \mathrm{V}_1=\frac{\mathrm{Q}}{\mathrm{C}_1}=\frac{120}{3}=40 \mathrm{~V} \
& \mathrm{~V}_2=\frac{\mathrm{Q}}{\mathrm{C}_2}=\frac{120}{2}=60 \mathrm{~V} \
	herefore \quad \mathrm{V}_2 & : \mathrm{V}_1=60: 40=3: 2
\end{aligned}$$

Answer

$$2: 3$$

Answer

$$3: 2$$

Answer

$$6: 5$$

Answer

$$5: 6$$

Electrostats & Capacitance

Two capacitors $\mathrm{C}_1=3 \mu \mathrm{F} and \mathrm{C}_2=2 \mu \mathrm{F} are connected in series across d.c. source of 100 \mathrm{~V}. The ratio of the potential across C_2 to C_1$ is

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