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7.Alternating Current
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An ac-circuit having supply voltage $E$ consists of a resistor of resistance $3\Omega$ and an inductor of reactance $4\Omega$ as shown in the figure. The voltage across the inductor at $t = \pi /\omega$ is.......$volts$
A
$2$
B
$10$
C
$4.8$
D
$0$
Solution
Given that, $R =3 \Omega$ and $X _{ L }=4 \Omega$
Total impedance, $Z =\sqrt{ R ^2+ X _{ L }^2}=\sqrt{3^2+4^2}=5 \Omega$
Voltage $E$ at $t=\mu \omega$ is
$E =10 \sin \omega(\mu \omega)=10 \sin \omega^2 \times 10^{-6}$
Value of $\omega^2 \times 10^{-6} \approx$ very low
therefore, $\sin \left(\omega^2 \times 10^{-6}\right) \approx 0$
Then current $I \rightarrow 0$
Hence, $iX L =$ Voltage across inductor
$V _{ L }=0$
Standard 12
Physics
