In a L-R growth circuit- inductance and resistance used are 1 H and 20 - respectively- If at t-0 50 ms- current in the circuit is 3-165A- then applied direct current emf is -

The correct option is A 100 V Time constant =120=50 ms so i= 0.633 i_max=0.633 ER⇒ E=3.165×200.633=100 V ...

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Standard XII

Physics

Ohm's Law in Scalar Form

In a L-R grow...

Question

In a L-R growth circuit, inductance and resistance used are $1 H$ and $20 Ω$ respectively. If at $t = 0$ $50 m s$ , current in the circuit is $3.165 A$ , then applied direct current emf is :

200 V

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100 V

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50 V

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Data is insufficient to find out the value

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Solution

The correct option is A $100 V$
Time constant $= \frac{1}{20} = 50 m s$
so $i = 0.633 i_{m a x} = 0.633 \frac{E}{R} \Rightarrow E = \frac{3.165 \times 20}{0.633} = 100 V$

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In a L-R growth circuit, inductance and resistance used are 1H and 20Ω respectively. If at t=0 50ms, current in the circuit is 3.165A, then applied direct current emf is :

The correct option is A 100VTime constant =120=50msso i=0.633imax=0.633ER⇒E=3.165×200.633=100V

In a L-R growth circuit, inductance and resistance used are $1 H$ and $20 Ω$ respectively. If at $t = 0$ $50 m s$ , current in the circuit is $3.165 A$ , then applied direct current emf is :

The correct option is A $100 V$
Time constant $= \frac{1}{20} = 50 m s$
so $i = 0.633 i_{m a x} = 0.633 \frac{E}{R} \Rightarrow E = \frac{3.165 \times 20}{0.633} = 100 V$