Rankers Physics

Capacitors: Practice Problem & Solution

A series combination of $n_{1}$ capacitors, each of value $C_{1}$, is charged by a source of potential difference $4V$. When another parallel combination of $n_{2}$ capacitors, each of value $C_{2}$, is charged by a source of potential difference $V$, it has the same (total) energy stored in it, as the first combination has. The value of $C_{2}$, in terms of $C_{1}$, is then: (2010 Pre)
$\frac{16C_{1}}{n_{1}n_{2}}$
$\frac{2C_{1}}{n_{1}n_{2}}$
$16\frac{n_{2}}{n_{1}}C_{1}$
$2\frac{n_{2}}{n_{1}}C_{1}$

Solution Explained:

To solve this problem, we apply the core principles of Capacitors. Understanding the underlying formula is key to arriving at the correct answer below:

Energy in series combination is $U_{1} = \frac{1}{2}(\frac{C_{1}}{n_{1}})(4V)^{2}$. Energy in parallel combination is $U_{2} = \frac{1}{2}(n_{2}C_{2})V^{2}$. Equating $U_{1}$ and $U_{2}$ gives $C_{2} = \frac{16C_{1}}{n_{1}n_{2}}$.

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Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. Two identical capacitors, have the same capacitance (C). One of them is charged to potential \(V_1\) and the other to \(V_2\). The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is :
2. A 400 pF capacitor is charged with a 100 V battery. After disconnecting the battery this capacitor is connected with another 400 pF capacitor. Then the energy loss is:
3. Choose the CORRECT statement :
4. Assertion (A): A parallel plate capacitor is charged to a potential difference of \( 100\text{V} \), and disconnected from the voltage source. A slab of dielectric is then slowly inserted between the plates. Compared to the energy before the slab was inserted, the energy stored in the capacitor with the dielectric is decreased. Reason (R): When we insert a dielectric between the plates of a capacitor, the induced charges tend to draw in the dielectric into the field (just as neutral objects are attracted by charged objects due to induction). We resist this force while slowly inserting the dielectric, and thus do negative work on the system, removing electrostatic energy from the system.  
5. Condenser A has a capacity of \(15 \mu\text{F}\) when it is filled with a medium of dielectric constant 15. Another condenser B has a capacity \(1 \mu\text{F}\) with air between the plates. Both are charged separately by a battery of \(100\text{ V}\). After charging, both are connected in parallel without the battery and the dielectric material being removed. The common potential now is :