SATHEE: Electro Magnetic Induction And Alternating Currents Question 31

Electro Magnetic Induction And Alternating Currents Question 31

Question: In a L-R circuit, the value of L is $(\frac{0.4}{π}) h e n r y$ and the value of R is 30 ohm. If in the circuit, an alternating e.m.f. of 200 volt at 50 cycles per sec is connected, the impedance of the circuit and current will be [MP PET 1996; DPMT 2003]

Options:

A) $11.4 Ω, 17.5 A$

B) $30.7 Ω, 6.5 A$

C) $40.4 Ω, 5 A$

D) $50 Ω, 4 A$

Show Answer

Answer:

Correct Answer: D

Solution:

$Z = \sqrt{R^{2} + X^{2}} = \sqrt{R^{2} + {(2 π f L)}^{2}}$ =

$\sqrt{{(30)}^{2} + {(2 π \times 50 \times \frac{0.4}{π})}^{2}} = \sqrt{900 + 1600} = 50 Ω$

$i = \frac{V}{Z} = \frac{200}{50} = 4 a m p e r e$

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Electro Magnetic Induction And Alternating Currents Question 31

Question: In a L-R circuit, the value of L is (0.4π) henry and the value of R is 30 ohm. If in the circuit, an alternating e.m.f. of 200 volt at 50 cycles per sec is connected, the impedance of the circuit and current will be [MP PET 1996; DPMT 2003]

Options:

Answer:

Solution:

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Question: In a L-R circuit, the value of L is $(\frac{0.4}{π}) h e n r y$ and the value of R is 30 ohm. If in the circuit, an alternating e.m.f. of 200 volt at 50 cycles per sec is connected, the impedance of the circuit and current will be [MP PET 1996; DPMT 2003]