A capacitor having capacity of 2-0mu F is charged to 200 V and then the plates of the capacitors are connected to a resistance wire- The heat produced in joule will be -2times 10-2-4times 10-2-4times 10-4-4times 10-10-

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Toppr Admin · Accepted Answer

Heat produce in a wire is equal to energy stored in capacitor-H-12CV2-12-xD7-2-xD7-10-x2212-6-xD7-200-2-10-x2212-6-xD7-200-xD7-200-4-xD7-10-x2212-2J-

A capacitor having capacity of $2.0 μ F$ is charged to $200 V$ and then the plates of the capacitors are connected to a resistance wire. The heat produced in joule will be :
$2 \times 10^{- 2}$
$4 \times 10^{- 2}$
$4 \times 10^{4}$
$4 \times 10^{10}$

Heat produce in a wire is equal to energy stored in capacitor.
$H = \frac{1}{2} C V^{2}$
$= \frac{1}{2} \times (2 \times 10^{- 6}) \times (200)^{2}$
$= 10^{- 6} \times 200 \times 200$
$= 4 \times 10^{- 2} J$ .

A capacitor having capacity of 2.0μF is charged to 200 V and then the plates of the capacitors are connected to a resistance wire. The heat produced in joule will be :2×10−24×10−24×1044×1010

Heat produce in a wire is equal to energy stored in capacitor.H=12CV2=12×(2×10−6)×(200)2=10−6×200×200=4×10−2J.

A capacitor having capacity of $2.0 μ F$ is charged to $200 V$ and then the plates of the capacitors are connected to a resistance wire. The heat produced in joule will be :
$2 \times 10^{- 2}$
$4 \times 10^{- 2}$
$4 \times 10^{4}$
$4 \times 10^{10}$

Heat produce in a wire is equal to energy stored in capacitor.
$H = \frac{1}{2} C V^{2}$
$= \frac{1}{2} \times (2 \times 10^{- 6}) \times (200)^{2}$
$= 10^{- 6} \times 200 \times 200$
$= 4 \times 10^{- 2} J$ .