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An 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 - Physics

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प्रश्न

An 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?

योग

उत्तर

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 = `V^2/R`
where R = resistance of electric bulb
`therefore R = V^2/P`

= `(110xx110)/55 = 220 Ω` 
If L is the inductance of the coil, then total reactance of the circuit (Z) is given by,
`Z = sqrt(R^2 + (omegaL)^2`
=`sqrt((220)^2 + (100piL)^2`      `( therefore omega = 2pif)`
Here, `omega` = angular frequency of the circuit
Now, current through the bulb, `I= V/Z`
∴ Voltage drop across the bulb, V = `V/ZxxR` 
As per question,
`110 = (220xx220)/sqrt((220)^2 + (100piL)^2`
`rArr  220xx2 = sqrt((220)^2 + (100piL)^2`
`rArr (220)^2 + (100piL)^2 = (440)^2`

`rArr 48400 + 10^4 pi^2L^2 = 193600`

`rArr 10^4 pi^2 L^2 = 193600 - 48400`

`rArr L^2 = 145200/(pi^2 xx 10^4)`

= 1.4726
`rArr L = 1.2135 ≅ 1.2 H`

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अध्याय 17: Alternating Current - Exercises [पृष्ठ ३३०]

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एचसी वर्मा Concepts of Physics Vol. 2 [English] Class 11 and 12
अध्याय 17 Alternating Current
Exercises | Q 13 | पृष्ठ ३३०

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