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GUJCET · Physics · Alternating Current
A coil has self inductance \(L=0.04 \mathrm{H}\) and resistance \(R=12 \mathrm{ohm}\). When it is connected to \(220 \mathrm{~V}, 50 \mathrm{~Hz}\) supply, what will be the current flowing through the coil ?
- A 12.7 A
- B 14.7 A
- C 11.7 A
- D 10.7 A
Answer & Solution
Correct Answer
(A) 12.7 A
Step-by-step Solution
Detailed explanation
(A) 12.7A
\(\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=2 \pi \nu \mathrm{~L}\)
\(X_{L}=2 \times 3.14 \times 50 \times 0.04\)
\(\mathrm{X}_{\mathrm{L}}=12.56 \Omega\)
\(\Rightarrow\) Impedance of circuit
\(
\begin{aligned}
& Z=\sqrt{R^{2}+\left(X_{L}\right)^{2}} \\
& Z=\sqrt{(12)^{2}+(12.56)^{2}} \\
& Z=17.37 \Omega
\end{aligned}
\)
\(\Rightarrow\) current passing through circuit
\(
\begin{aligned}
& I=\frac{V}{Z} \\
& I=\frac{220}{17.37} \\
& I \approx 12.7 \mathrm{~A}
\end{aligned}
\)
\(\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=2 \pi \nu \mathrm{~L}\)
\(X_{L}=2 \times 3.14 \times 50 \times 0.04\)
\(\mathrm{X}_{\mathrm{L}}=12.56 \Omega\)
\(\Rightarrow\) Impedance of circuit
\(
\begin{aligned}
& Z=\sqrt{R^{2}+\left(X_{L}\right)^{2}} \\
& Z=\sqrt{(12)^{2}+(12.56)^{2}} \\
& Z=17.37 \Omega
\end{aligned}
\)
\(\Rightarrow\) current passing through circuit
\(
\begin{aligned}
& I=\frac{V}{Z} \\
& I=\frac{220}{17.37} \\
& I \approx 12.7 \mathrm{~A}
\end{aligned}
\)
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