SATHEE: Electro Magnetic Induction And Alternating Currents Question 348

Electro Magnetic Induction And Alternating Currents Question 348

Question: A 50 volt potential difference is suddenly applied to a coil with $L = 5 \times 10^{- 3}$ henry and $R = 180, o h m$ . The rate of increase of current after 0.001 second is [MP PET 1994]

Options:

A) 27.3 amp/sec

B) 27.8 amp/sec

C) 2.73 amp/sec

D) None of the above

Show Answer

Answer:

Correct Answer: D

Solution:

The rate of increase of current

$= \frac{d i}{d t} = \frac{d}{d t} i_{0} (1 - e^{- R t / L}) = \frac{d}{d t} i_{0} - \frac{d}{d t} i_{0} e^{- R t / L}$

$= 0 - i_{0} e^{- R t / L} . \frac{d}{d t} (- \frac{R t}{L}) = i_{0} \frac{R}{L} e^{- R t / L}$

$= \frac{50}{180} \times \frac{180}{5 \times 10^{- 3}} \times e^{- (180 \times 0.001) / (5 \times 10^{- 3})}$ $= 10^{4} \times e^{- 36} A / \sec$

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Electro Magnetic Induction And Alternating Currents Question 348

Question: A 50 volt potential difference is suddenly applied to a coil with L=5×10−3 henry and R=180,ohm . The rate of increase of current after 0.001 second is [MP PET 1994]

Options:

Answer:

Solution:

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Question: A 50 volt potential difference is suddenly applied to a coil with $L = 5 \times 10^{- 3}$ henry and $R = 180, o h m$ . The rate of increase of current after 0.001 second is [MP PET 1994]