Question

The intensity of a lightbulb with a resistance of 110 ohms is controlled by connecting it in series with an inductor whose inductance can be varied from L=0 to L =Lmax. This "light dimmer" circuit is connected to an ac generator with frequency of 60hz. And an rms voltage of 120V. P=130 w The inductor is now adjusted to L=Lmax. The average power dissipated in the lightbulb is one-fourth of 130 W. What is the value of Lmax?

Answer 1

Answer on Question #79555, Physics / Other

The intensity of a lightbulb with a resistance of 110 ohms is controlled by connecting it in series with an inductor whose inductance can be varied from $L=0$ to $L=L_{\max}$. This "light dimmer" circuit is connected to an ac generator with frequency of $60~\mathrm{Hz}$ and an rms voltage of $120\mathrm{V}$. $P=130\mathrm{w}$ The inductor is now adjusted to $L=L_{\max}$. The average power dissipated in the lightbulb is one-fourth of $130\mathrm{W}$. What is the value of $L_{\max}$?

Solution:

The rms current flowing in the circuit:

$$I = \frac {V}{Z}$$

where, $Z$ is impedance of the circuit.

$$Z = \sqrt {R ^ {2} + \omega^ {2} L _ {\max} ^ {2}}$$

The average power dissipated of the circuit is

$$P _ {a v} = V I \cos \varphi$$

The power factor of the circuit

$$\cos \varphi = \frac {R}{Z}$$ $$P _ {a v} = \frac {V ^ {2} R}{Z ^ {2}} = \frac {1}{4} \times 1 3 0 W$$ $$\frac {V ^ {2} R}{R ^ {2} + \omega^ {2} L _ {\max} ^ {2}} = 3 2. 5 W$$

So,

$$L _ {m a x} = \frac {1}{\omega^ {2}} \left(\frac {V ^ {2} R}{3 2 . 5} - R ^ {2}\right) = \frac {1}{4 \pi^ {2} f ^ {2}} \left(\frac {V ^ {2} R}{3 2 . 5} - R ^ {2}\right)$$ $$L _ {m a x} = \frac {1}{4 \pi^ {2} (6 0) ^ {2}} \left(\frac {1 2 0 ^ {2} \times 1 1 0}{3 2 . 5} - 1 1 0 ^ {2}\right) = 0. 2 5 8 H$$

Answer: 0.258 H

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