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  4. Capacitor C3 in the circuit is a variable capacitor (its capacitance can be varied). Graph is plotted between potential difference V1 (across capacitor C1 ) versus C3. Electric potential V1 approaches on asymptote of 10 V as C3 arrow ∞. Then C1 / C2 has value

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Q. Capacitor $C_{3}$ in the circuit is a variable capacitor (its capacitance can be varied). Graph is plotted between potential difference $V_{1}$ (across capacitor $C_{1}$ ) versus $C_{3}$. Electric potential $V_{1}$ approaches on asymptote of $10\, V$ as $C_{3} \rightarrow \infty$. Then $C_{1} / C_{2}$ has valuePhysics Question Image

Electrostatic Potential and Capacitance

Solution:

For $C_{3} \rightarrow \infty$ :
image
$\Rightarrow V=10\, V\,\,\,...(i)$
For $C_3 = 0$
image
Now, $V=\frac{Q}{C}$
$\therefore \frac{2}{8}=\frac{C_{2}}{C_{1}}$
$C_{1}=4 C_{2}$