JEE MAIN - Physics (2018 - 15th April Morning Slot - No. 24)
An ideal capacitor of capacitance $$0.2\,\mu F$$ is charged to a potential difference of $$10$$ $$V.$$ The charging battery is then disconnected. The capacitor is then connected to an ideal inductor of self inductance $$0.5$$ $$mH.$$ The current at a time when the potential difference across the capacitor is $$5$$ $$V,$$ is :
$$0.34\,\,A$$
$$0.25\,\,A$$
$$0.17\,\,A$$
$$0.15\,\,A$$
Explanation
Capacitance, C = 0.2 $$\mu $$F = 0.2 $$ \times $$ 10$$-$$6 F
Inductance, L = 0.5 m H = 0.5 $$ \times $$ 10$$-$$3 H
Let, current = I.
Using energy conservation,
UE + 0 = UE' + Ub'
$$ \Rightarrow $$$$\,\,\,\,$$ $${1 \over 2}$$ cv2 + 0 = $${1 \over 2}$$ c$$v_1^2$$ + $${1 \over 2}$$LI2
$$ \Rightarrow $$$$\,\,\,\,$$ $${1 \over 2}$$ $$ \times $$ 0.2 $$ \times $$ 10$$-$$6 $$ \times $$ 102
= $${1 \over 2} \times $$ 0.2 $$ \times $$ 10$$-$$6 $$ \times $$ 52 + $${1 \over 2}$$ $$ \times $$ 0.5 $$ \times $$ 10$$-$$3 $$ \times $$ I2
By solving this,
I = $$\sqrt 3 \times $$ 10$$-$$1 A
= 0.17 A.
Inductance, L = 0.5 m H = 0.5 $$ \times $$ 10$$-$$3 H
Let, current = I.
Using energy conservation,
UE + 0 = UE' + Ub'
$$ \Rightarrow $$$$\,\,\,\,$$ $${1 \over 2}$$ cv2 + 0 = $${1 \over 2}$$ c$$v_1^2$$ + $${1 \over 2}$$LI2
$$ \Rightarrow $$$$\,\,\,\,$$ $${1 \over 2}$$ $$ \times $$ 0.2 $$ \times $$ 10$$-$$6 $$ \times $$ 102
= $${1 \over 2} \times $$ 0.2 $$ \times $$ 10$$-$$6 $$ \times $$ 52 + $${1 \over 2}$$ $$ \times $$ 0.5 $$ \times $$ 10$$-$$3 $$ \times $$ I2
By solving this,
I = $$\sqrt 3 \times $$ 10$$-$$1 A
= 0.17 A.
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