JEE MAIN - Physics (2021 - 26th February Evening Shift - No. 3)
Find the peak current and resonant frequency of the following circuit (as shown in figure).
_26th_February_Evening_Shift_en_3_1.png)
2 A and 100 Hz
2 A and 50 Hz
0.2 A and 100 Hz
0.2 A and 50 Hz
Explanation
We know, z = $$\sqrt {{{({x_L} - {x_C})}^2} + {R^2}} $$
$${x_L} = {\omega _L} = 100 \times 100 \times {10^{ - 3}} = 10\Omega $$
$${x_C} = {1 \over {{\omega _C}}} = {1 \over {100 \times 100 \times {{10}^{ - 6}}}} = 10\Omega $$
$$ \therefore $$ $$z = \sqrt {{{(10 - 100)}^2} + {R^2}} = \sqrt {{{90}^2} + {{120}^2}} $$
$$ = 30 \times 5 = 150\Omega $$
$${i_{peak}} = {{\Delta v} \over z} = {{30} \over {150}} = {1 \over 5} amp = 0.2$$ amp
For resonant frequency,
$$ \omega L = {1 \over {\omega C}} \Rightarrow {\omega ^2} = {1 \over {LC}} \Rightarrow \omega = {1 \over {\sqrt {LC} }}$$
& $$f = {1 \over {2\pi \sqrt {LC} }} = {1 \over {2\pi \sqrt {100 \times {{10}^{ - 3}} \times 100 \times {{10}^{ - 6}}} }}$$
$$ = {{100\sqrt {10} } \over {2\pi }} = {{100\pi } \over {2\pi }} = 50$$ Hz
as $$\sqrt {10} \approx \pi $$
$${x_L} = {\omega _L} = 100 \times 100 \times {10^{ - 3}} = 10\Omega $$
$${x_C} = {1 \over {{\omega _C}}} = {1 \over {100 \times 100 \times {{10}^{ - 6}}}} = 10\Omega $$
$$ \therefore $$ $$z = \sqrt {{{(10 - 100)}^2} + {R^2}} = \sqrt {{{90}^2} + {{120}^2}} $$
$$ = 30 \times 5 = 150\Omega $$
$${i_{peak}} = {{\Delta v} \over z} = {{30} \over {150}} = {1 \over 5} amp = 0.2$$ amp
For resonant frequency,
$$ \omega L = {1 \over {\omega C}} \Rightarrow {\omega ^2} = {1 \over {LC}} \Rightarrow \omega = {1 \over {\sqrt {LC} }}$$
& $$f = {1 \over {2\pi \sqrt {LC} }} = {1 \over {2\pi \sqrt {100 \times {{10}^{ - 3}} \times 100 \times {{10}^{ - 6}}} }}$$
$$ = {{100\sqrt {10} } \over {2\pi }} = {{100\pi } \over {2\pi }} = 50$$ Hz
as $$\sqrt {10} \approx \pi $$
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