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. Assertion (A): It is not possible to make a spherical conductor of capacitor one farad. Reason (R): It is possible for earth as its radius is \( 6400 \text{ km} \).  
2. The energy stored and potential $V$ is in a capacitor of capacity $C$ given by (1996)
3. A charge of \( + 2.0 \times 10^{-8}\text{ C} \) is placed on the positive plate and a charge of \( -1.0 \times 10^{-8}\text{ C} \) on the negative plate of a parallel-plate capacitor of capacitance \( 1.2 \times 10^{-3}\ \mu\text{F} \). Calculate the potential difference developed between the plates.
4. Assertion (A): In parallel plate capacitor separation 'd' should be smaller than the linear dimension of the plates \( d^2 << A\). Reason (R): For \( d^2 << A \) a fringing effect can be ignored in the region sufficiently far from the edge.  
5. A parallel plate capacitor has a uniform electric field \(\vec{E}\) in the space between the plates. If the distance between the plates is \(d\) and the area of each plate is \(A\), the energy stored in the capacitor is (\(\varepsilon_0\) = permittivity of free space)