KCET · Physics · Electromagnetic Induction
An electric bulb has a rated power of \(50 \mathrm{~W}\) at \(100 \mathrm{~V}\). If it is used on an AC source \(200 \mathrm{~V}, 50\) \(\mathrm{Hz}\), a choke has to be used in series with it. This choke should have an inductance of
- A \(0.1 \mathrm{mH}\)
- B \(1 \mathrm{mH}\)
- C \(0.1 \mathrm{H}\)
- D \(1.1 \mathrm{H}\)
Answer & Solution
Correct Answer
(D) \(1.1 \mathrm{H}\)
Step-by-step Solution
Detailed explanation
Resistance of bulb
\(R=\frac{V^{2}}{P}=\frac{(100)^{2}}{50}=200 \Omega\)
Current through bulb \((I)=\frac{V}{R}\)
\(=\frac{100}{200}=0.5 \mathrm{~A}\)
In a circuit containing inductive reactance \(\left(X_{L}\right)\) and resistance \((R)\), impedance \((Z)\) of the circuit is
\(Z=\sqrt{R^{2}+\omega^{2} L^{2}}\)
Here, \(Z=\frac{200}{0.5}=400 \Omega\)
Now, \(X_{L}^{2}=Z^{2}-R^{2}=(400)^{2}-(200)^{2}\)
\(\begin{aligned}(2 \pi f L)^{2} &=12 \times 10^{4} \\L &=\frac{2 \sqrt{3} \times 100}{2 \pi \times 50}=\frac{2 \sqrt{3}}{\pi}=1.1 \mathrm{H}\end{aligned}\)
\(R=\frac{V^{2}}{P}=\frac{(100)^{2}}{50}=200 \Omega\)
Current through bulb \((I)=\frac{V}{R}\)
\(=\frac{100}{200}=0.5 \mathrm{~A}\)
In a circuit containing inductive reactance \(\left(X_{L}\right)\) and resistance \((R)\), impedance \((Z)\) of the circuit is
\(Z=\sqrt{R^{2}+\omega^{2} L^{2}}\)
Here, \(Z=\frac{200}{0.5}=400 \Omega\)
Now, \(X_{L}^{2}=Z^{2}-R^{2}=(400)^{2}-(200)^{2}\)
\(\begin{aligned}(2 \pi f L)^{2} &=12 \times 10^{4} \\L &=\frac{2 \sqrt{3} \times 100}{2 \pi \times 50}=\frac{2 \sqrt{3}}{\pi}=1.1 \mathrm{H}\end{aligned}\)
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