In an $L R$ circuit, the value of $L$ is $\frac{0.4}{π} H$ and the value $R$ is $30 Ω$ . If in the circuit, an alternating $e m f$ of $200 V$ $r m s$ value at $50 cycles/s$ is connected, the peak current of the circuit is $k \sqrt{2} A$ .The value of $k$ is (integer only)

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Solution

Impedence for $L R$ circuit is given by,

$Z = \sqrt{R^{2} + X_{L}^{2}} = \sqrt{R^{2} + ω^{2} L^{2}}$

As, $f = 50 Hz$ $\Rightarrow ω = 2 π f = 100 π$

$∴ X_{L} = ω L = 100 π \frac{(0.4)}{π} = 40 Ω$

So, $Z = \sqrt{30^{2} + 40^{2}} = 50 Ω$

$I_{r m s} = \frac{V_{r m s}}{Z} = \frac{200}{50} = 4 A$

$∴ I_{p e a k} = I_{r m s} \sqrt{2} = 4 \sqrt{2}$

Given, $I_{p e a k} = k \sqrt{2}$

$\Rightarrow k = 4$

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In an LR circuit, the value of L is 0.4π H and the value R is 30 Ω. If in the circuit, an alternating emf of 200 V rms value at 50 cycles/s is connected, the peak current of the circuit is k√2 A .The value of k is (integer only) ​​​

Impedence for LR circuit is given by, Z=√R2+X2L=√R2+ω2L2 As, f=50 Hz ⇒ω=2πf=100π ∴XL=ωL=100π(0.4)π=40 Ω So, Z=√302+402=50 Ω Irms=VrmsZ=20050=4 A ∴Ipeak=Irms√2=4√2 Given, Ipeak=k√2 ⇒k=4

In an $L R$ circuit, the value of $L$ is $\frac{0.4}{π} H$ and the value $R$ is $30 Ω$ . If in the circuit, an alternating $e m f$ of $200 V$ $r m s$ value at $50 cycles/s$ is connected, the peak current of the circuit is $k \sqrt{2} A$ .The value of $k$ is (integer only)