JEE MAIN - Physics (2020 - 9th January Evening Slot - No. 18)
In LC circuit the inductance L = 40 mH and
capacitance C = 100 $$\mu $$F. If a voltage
V(t) = 10sin(314t) is applied to the circuit, the
current in the circuit is given as :
capacitance C = 100 $$\mu $$F. If a voltage
V(t) = 10sin(314t) is applied to the circuit, the
current in the circuit is given as :
0.52 cos 314 t
5.2 cos 314 t
0.52 sin 314 t
10 cos 314 t
Explanation
_9th_January_Evening_Slot_en_18_1.png)
Z = xC – xL
= $${1 \over {\omega C}} - \omega L$$
= $${1 \over {314 \times 100 \times {{10}^{ - 6}}}} - 314 \times 40 \times {10^{ - 3}}$$
= 19.28 $$\Omega $$
As Vm = ImZ
$$ \Rightarrow $$ 10 = Im $$ \times $$ 19.28
$$ \Rightarrow $$ Im = $${{10} \over {19.28}}$$ = 0.52 A
$$ \therefore $$ I = 0.52 sin(314t + $${\pi \over 2}$$)
= 0.52 cos(314t)
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