A condenser having a capacity 2.0 μF is charged to 200 V and then the plates of the capacitor are connected to a resistance wire. What will be the heat produced in joules ? In this case heat produced is equapacitor

Question

A condenser having a capacity 2.0 μF is charged to 200 V and then the plates of the capacitor are connected to a resistance wire. What will be the heat produced in joules ? In this case heat produced is equapacitor (A) 1 × 10-2 J (B) 2 × 10-2 J (C) 3 × 10-2 J (D) 4 × 10-2 J

Tardigrade Admin · Accepted Answer

Here, H =(1/2) CV2 = (1/2) × 2 × 10-6 × 2002 = 10-6 × 40000 = 4 × 10-2 joule.

Q. A condenser having a capacity $2.0$ $μ$ F is charged to $200 V$ and then the plates of the capacitor are connected to a resistance wire. What will be the heat produced in joules ?
In this case heat produced is equapacitor

$1 \times 1 0^{- 2} J$

$2 \times 1 0^{- 2} J$

$3 \times 1 0^{- 2} J$

$4 \times 1 0^{- 2} J$

Here, H = $\frac{1}{2} C V^{2} = \frac{1}{2} \times 2 \times 1 0^{- 6} \times 20 0^{2}$
= $1 0^{- 6} \times 40000$
= $4 \times 1 0^{- 2}$ joule.

Q. A condenser having a capacity 2.0 μF is charged to 200V and then the plates of the capacitor are connected to a resistance wire. What will be the heat produced in joules ? In this case heat produced is equapacitor

1×10−2J

2×10−2J

3×10−2J

4×10−2J