SATHEE: Electro Magnetic Induction And Alternating Currents Question 131

Electro Magnetic Induction And Alternating Currents Question 131

Question: One 10 V, 60 W bulb is to be connected to 100 V line. The required induction coil has self inductance of value $(f = 50, H z)$ [RPET 1997]

Options:

A) 0.052 H

B) 2.42 H

C) 16.2 mH

D) 1.62 mH

Show Answer

Answer:

Correct Answer: A

Solution:

Current through the bulb $i = \frac{P}{V} = \frac{60}{10} = 6 A$ $V = \sqrt{V_{R}^{2} + V_{L}^{2}}$

${(100)}^{2} = {(10)}^{2} + V_{L}^{2}$

$\Rightarrow V_{L} = 99.5 V o l t$

Also $V_{L} = i X_{L} = i \times (2 π ν L)$

$\Rightarrow, 99.5 = 6 \times 2 \times 3.14 \times 50 \times L$

$\Rightarrow L = 0.052 H$

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Electro Magnetic Induction And Alternating Currents Question 131

Question: One 10 V, 60 W bulb is to be connected to 100 V line. The required induction coil has self inductance of value (f=50,Hz) [RPET 1997]

Options:

Answer:

Solution:

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Question: One 10 V, 60 W bulb is to be connected to 100 V line. The required induction coil has self inductance of value $(f = 50, H z)$ [RPET 1997]