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Standard XII

Physics

Alternating Current Applied to an Inductor

In an LCR cir...

Question

In an LCR circuit, the value of L is $\frac{0.4}{π}$ henry and the value of R is 30 $Ω$ . If in the circuit, an alternating e.m.f. of 200 V at 50 cycles per second is connected the impedance of the circuit and current will be

11.4

Ω

, 17.5 A

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30.7

Ω

, 6.5 A

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40.4

Ω

, 5 A

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Ω

, 4 A

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Solution

The correct option is D 50 $Ω$ , 4 A
$r e a c t a n e = ω L = 2 π \times 50 \times \frac{.4}{π} = 40 i m p e d a n c e = \sqrt{{r e s i s t a n c e}^{2} + {r e a c t a n c e}^{2}} = \sqrt{30^{2} + 40^{2}} = 50 i = \frac{V}{i m p e d a n c e} = \frac{200}{50} = 4 a m p$

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Similar questions

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In an LCR circuit, the value of L is 0.4π henry and the value of R is 30Ω. If in the circuit, an alternating e.m.f. of 200 V at 50 cycles per second is connected the impedance of the circuit and current will be

The correct option is D 50 Ω, 4 Areactane=ωL=2π×50×.4π=40impedance=√resistance2+reactance2=√302+402=50i=Vimpedance=20050=4amp

In an LCR circuit, the value of L is $\frac{0.4}{π}$ henry and the value of R is 30 $Ω$ . If in the circuit, an alternating e.m.f. of 200 V at 50 cycles per second is connected the impedance of the circuit and current will be

The correct option is D 50 $Ω$ , 4 A
$r e a c t a n e = ω L = 2 π \times 50 \times \frac{.4}{π} = 40 i m p e d a n c e = \sqrt{{r e s i s t a n c e}^{2} + {r e a c t a n c e}^{2}} = \sqrt{30^{2} + 40^{2}} = 50 i = \frac{V}{i m p e d a n c e} = \frac{200}{50} = 4 a m p$