| Direction: Q.46 to Q.50 |
| A capacitor is a device to store energy. The process of charging up a capacitor involves the transferring of electric charges from its one place to another. This work done in charging the capacitor is stored as its electrical potential energy. |
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| If q is the charge and V is the potential difference across a capacitor at any instant during its charging, then small work done in storing an additional small charge dq against the repulsion of charge q already stored on it is \[dW=V.dq=\left( q/C \right)dq.\] |
| Read the given passage carefully and give the answer of the following questions. |
A) \[3\mu J\]
B) \[24\mu J\]
C) \[30\mu J\]
D) \[108\,\mu J\]
Correct Answer: B
Solution :
(b) \[24\mu J\] As, \[{{C}_{1}}=2\mu F,\,{{C}_{2}}=4\mu F\] In series combination, the equivalent capacitance will be, \[C=\frac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}=\left( \frac{2\times 4}{2+4} \right)\mu F=\frac{4}{3}\mu F\] Potential difference applied, V = 6 V Energy stored in the system, \[U=\frac{1}{2}C{{V}^{2}}\] \[=\frac{1}{2}C{{V}^{2}}=\frac{1}{2}\times \left( 10\times {{10}^{-6}} \right){{\left( 10 \right)}^{2}}=500\mu J\]
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