- 1(current)
Q. No.Two identical capacitors \(C_{1}\) and \(C_{2}\) of equal capacitance are connected as shown in the circuit. Terminals \(a\) and \(b\) of the key \(k\) are connected to charge capacitor \(C_{1}\) using a battery of emf \(V\) volt. Now disconnecting \(a\) and \(b\) terminals, terminals \(b\) and \(c\) are connected. Due to this, what will be the percentage loss of energy?
1. \(75\%\)
2. \(0\%\)
3. \(50\%\)
4. \(25\%\)
| (A) | The charge stored in it increases. |
| (B) | The energy stored in it decreases. |
| (C) | Its capacitance increases. |
| (D) | The ratio of charge to its potential remains the same. |
| (E) | The product of charge and voltage increases. |
| 1. | (A), (C) and (E) only |
| 2. | (B), (D) and (E) only |
| 3. | (A), (B) and (C) only |
| 4. | (A), (B) and (E) only |
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:
\(\left ( \varepsilon_{0} = \text{permittivity of free space} \right )\)
| 1. | \(\dfrac{1}{2}\varepsilon_0 E^2 Ad\) | 2. | \(\dfrac{E^2 Ad}{\varepsilon_0}\) |
| 3. | \(\dfrac{1}{2}\varepsilon_0 E^2 \) | 4. | \(\varepsilon_0 EAd\) |
In the circuit shown in the figure, the energy stored in \(6~\mu\text{F}\) capacitor will be:
| 1. | \(48 \times10^{-6}~\text{J}\) | 2. | \(32 \times10^{-6}~\text{J}\) |
| 3. | \(96 \times10^{-6}~\text{J}\) | 4. | \(24 \times10^{-6}~\text{J}\) |
| 1. | \(\dfrac{1}{2} \varepsilon_{o} \dfrac{V^{2}}{d^{2}}\) | 2. | \(\dfrac{1}{2 \varepsilon_{o}} \dfrac{V^{2}}{d^{2}}\) |
| 3. | \(\dfrac{1}{2} C V^{2}\) | 4. | \(\dfrac{Q^{2}}{2 C}\) |
1. \(1.7\times10^{-6}~\text J\)
2. \(2.12\times10^{-6}~\text J\)
3. \(2.55\times10^{-6}~\text J\)
4. \(1.66\times10^{-6}~\text J\)
- 1(current)
Q. No.
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