JEE Advance - Physics (2021 - Paper 2 Online - No. 12)
In a circuit, a metal filament lamp is connected in series with a capacitor of capacitance C $$\mu$$F across a 200 V, 50 Hz supply. The power consumed by the lamp is 500 W while the voltage drop across it is 100 V. Assume that there is no inductive load in the circuit. Take rms values of the voltages. The magnitude of the phase-angle (in degrees) between the current and the supply voltage is $$\varphi $$. Assume, $$\pi$$$$\sqrt 3 $$ $$ \approx $$ 5.
The value of $$\varphi$$ is ____________.
The value of $$\varphi$$ is ____________.
Answer
60
Explanation
For lamp,
$$p = {{{V^2}} \over R}$$
$$R = {{100 \times 100} \over {500}} = 20\Omega $$
$$i = {V \over R} = {{100} \over {20}} = 5A \Rightarrow i = {V \over {\sqrt {{R^2} + X_C^2} }}$$
$$ \Rightarrow 5 = {{200} \over {\sqrt {{{(20)}^2} + {{\left( {{1 \over {2\pi \times 50 \times C}}} \right)}^2}} }}$$
$$ \Rightarrow 400 + {\left( {{1 \over {100\pi C}}} \right)^2} = 1600$$
$$ \Rightarrow {\left( {{1 \over {100\pi C}}} \right)^2} = 1200 \Rightarrow {1 \over {{\pi ^2}{C^2}}} = 1200 \times {10^4}$$
$$ \Rightarrow {(\pi C)^2} = {1 \over {12}} \times {10^{ - 6}}$$
$$ \Rightarrow {C^2} = {1 \over {12{\pi ^2}}} \times {10^{ - 6}} $$
$$ \Rightarrow $$ $$ C = {1 \over {\sqrt {12} \pi}} \times {10^{ - 3}} = 100 \times {10^{ - 6}}F = 100\mu F$$
$$V = \sqrt {V_C^2 + V_R^2} $$
$$200 = \sqrt {V_C^2 + {{100}^2}} $$
$$ \Rightarrow {V_C} = 100\sqrt 3 $$
$$\tan \phi = {{{X_C}} \over R} = {{{V_C}} \over {{V_R}}} = {{100\sqrt 3 } \over {100}} = \sqrt 3 $$
$$\phi = 60^\circ $$
$$p = {{{V^2}} \over R}$$
$$R = {{100 \times 100} \over {500}} = 20\Omega $$
$$i = {V \over R} = {{100} \over {20}} = 5A \Rightarrow i = {V \over {\sqrt {{R^2} + X_C^2} }}$$
$$ \Rightarrow 5 = {{200} \over {\sqrt {{{(20)}^2} + {{\left( {{1 \over {2\pi \times 50 \times C}}} \right)}^2}} }}$$
$$ \Rightarrow 400 + {\left( {{1 \over {100\pi C}}} \right)^2} = 1600$$
$$ \Rightarrow {\left( {{1 \over {100\pi C}}} \right)^2} = 1200 \Rightarrow {1 \over {{\pi ^2}{C^2}}} = 1200 \times {10^4}$$
$$ \Rightarrow {(\pi C)^2} = {1 \over {12}} \times {10^{ - 6}}$$
$$ \Rightarrow {C^2} = {1 \over {12{\pi ^2}}} \times {10^{ - 6}} $$
$$ \Rightarrow $$ $$ C = {1 \over {\sqrt {12} \pi}} \times {10^{ - 3}} = 100 \times {10^{ - 6}}F = 100\mu F$$
$$V = \sqrt {V_C^2 + V_R^2} $$
$$200 = \sqrt {V_C^2 + {{100}^2}} $$
$$ \Rightarrow {V_C} = 100\sqrt 3 $$
$$\tan \phi = {{{X_C}} \over R} = {{{V_C}} \over {{V_R}}} = {{100\sqrt 3 } \over {100}} = \sqrt 3 $$
$$\phi = 60^\circ $$
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