JEE MAIN - Physics (2020 - 2nd September Morning Slot - No. 19)
A 5 $$\mu $$F capacitor is charged fully by a 220 V
supply. It is then disconnected from the supply
and is connected in series to another
uncharged 2.5 $$\mu $$F capacitor. If the energy
change during the charge redistribution is
$${X \over {100}}J$$ then value of X to the nearest integer is
_____.
Answer
4
Explanation
ui = $$\frac{1}{2} $$ $$ \times $$ 5 $$ \times $$ 10-6$$ \times $$220
Final common potential
= $$\frac{220\times 5+0\times 2.5}{5+2.5} $$ = 220 $$ \times $$ $$\frac{2}{3} $$
uf = $$\frac{1}{2} $$ $$ \times $$ (5 + 2.5)$$ \times $$10-6 $$ \times $$ $$\left( 220\times \frac{2}{3} \right)^{2} $$
$$\Delta $$u = ui - uf
$$ \Rightarrow $$ $$\Delta $$u = –403.33 × 10–4
$$ \Rightarrow $$ –403.33 × 10–4 = $${X \over {100}}J$$
$$ \Rightarrow $$ X = -4.03
Value of X is approximate 4
Final common potential
= $$\frac{220\times 5+0\times 2.5}{5+2.5} $$ = 220 $$ \times $$ $$\frac{2}{3} $$
uf = $$\frac{1}{2} $$ $$ \times $$ (5 + 2.5)$$ \times $$10-6 $$ \times $$ $$\left( 220\times \frac{2}{3} \right)^{2} $$
$$\Delta $$u = ui - uf
$$ \Rightarrow $$ $$\Delta $$u = –403.33 × 10–4
$$ \Rightarrow $$ –403.33 × 10–4 = $${X \over {100}}J$$
$$ \Rightarrow $$ X = -4.03
Value of X is approximate 4
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