NEET-XII-Physics
39: Alternating Current
- #13An electric bulb is designed to consume 55 W when operated at 110 volts. It is connected to a 220 V, 50 Hz line through a choke coil in series. What should be the inductance of the coil for which the bulb gets correct voltage?Ans : Power consumed by the electric bulb, P = 55 W
Voltage at which the bulb is operated, V= 110 V
Voltage of the line, V = 220 V
Frequency of the source, v = 50 Hz
P =`` \frac{{V}^{2}}{R}``,
where R = resistance of electric bulb
`` \therefore R=\frac{{V}^{2}}{P}``
`` =\frac{110\times 110}{55}=220\,\mathrm{\,\Omega \,}``
If L is the inductance of the coil, then total reactance of the circuit (Z) is given by,
Z = `` \sqrt{{R}^{2}+{\left(\omega L\right)}^{2}}``
=`` \sqrt{{\left(220\right)}^{2}+{\left(100\,\mathrm{\,\pi \,}L\right)}^{2}}`` `` \left(\because \omega =2\,\mathrm{\,\pi \,}f\right)``
Here, `` \omega `` = angular frequency of the circuit
Now, current through the bulb, `` I=\frac{V}{Z}``
`` \therefore `` Voltage drop across the bulb, V = `` \frac{V}{Z}\times R``
As per question,
`` 110=\frac{220\times 220}{\sqrt{{\left(220\right)}^{2}+{\left(100\,\mathrm{\,\pi \,}L\right)}^{2}}}``
`` 110=\frac{{\left(220\right)}^{2}}{\sqrt{{\left(220\right)}^{2}+{\left(100\,\mathrm{\,\pi \,}L\right)}^{2}}}``
`` \Rightarrow 220\times 2=\sqrt{{\left(220\right)}^{2}+{\left(100\,\mathrm{\,\pi \,}L\right)}^{2}}``
`` \Rightarrow {\left(220\right)}^{2}+{\left(100\,\mathrm{\,\pi \,}L\right)}^{2}={\left(440\right)}^{2}``
`` \Rightarrow 48400+{10}^{4}{\,\mathrm{\,\pi \,}}^{2}{L}^{2}=193600``
`` \Rightarrow {10}^{4}{\,\mathrm{\,\pi \,}}^{2}{L}^{2}=193600-48400``
`` \Rightarrow {L}^{2}=\frac{145200}{{\,\mathrm{\,\pi \,}}^{2}\times {10}^{4}}``
`` =1.4726``
`` \Rightarrow L=1.2135\cong 1.2\,\mathrm{\,H\,}``
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