You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Magnetic Flux and Faraday's Law

The network s...

Question

The network shown in the figure is part of a complete circuit. If at a certain instant, the current $I$ is $5 A$ and it is decreasing at a rate of $10^{3} A s^{- 1}$ then $V_{B} - V_{A}$ equals :

20 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

15 V

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

10 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $15 V$
We will write voltage drop equation as we go from A to B.
$V_{A} - i R + V + L \frac{d i}{d t} = V_{B}$
$V_{A} - 5 \times 1 + 15 - 5 \times 10^{- 3} \times (- 10^{3}) = V_{B}$
$V_{B} - V_{A} = 15$

Suggest Corrections

Similar questions

The network shown in the figure is part of a complete circuit. If at a certain instant, the current I is 5 A and it is decreasing at a rate of $10^{3} A s^{- 1}$ then $V_{B} - V_{A}$ equals

Q. The network shown in the figure is part of a complete circuit. If at a certain instant, the current

i

5 A

and decreasing at a rate of

10^{3} {A s}^{- 1}

then potential difference between

A

and

B

(

V_{B} - V_{A}

) is equal to

Q. The network shown in the figure is part of a circuit. If at a certain instant, the current

I

5 A

flowing through the circuit part and it is decreasing at a rate of

10^{3} {As}^{- 1}

then

V_{B} - V_{A}

equals

Q. The network shown in the figure is part of a complete circuit. If at a certain instant, the current I is

5

A and it is decreasing at a rate of

10^{3}

A s^{- 1}

then

V_{B} - V_{A}

equals.

The figure shows a part of a complete circuit. The potential difference $V_{B} - V_{A}$ when the current I is 5A and is decreasing at a rate of $10^{3}$ A $s^{- 1}$ is given by

Join BYJU'S Learning Program

The network shown in the figure is part of a complete circuit. If at a certain instant, the current I is 5A and it is decreasing at a rate of 103As−1 then VB−VA equals :

The correct option is B 15VWe will write voltage drop equation as we go from A to B.VA−iR+V+Ldidt=VBVA−5×1+15−5×10−3×(−103)=VBVB−VA=15

The network shown in the figure is part of a complete circuit. If at a certain instant, the current $I$ is $5 A$ and it is decreasing at a rate of $10^{3} A s^{- 1}$ then $V_{B} - V_{A}$ equals :

The correct option is B $15 V$
We will write voltage drop equation as we go from A to B.
$V_{A} - i R + V + L \frac{d i}{d t} = V_{B}$
$V_{A} - 5 \times 1 + 15 - 5 \times 10^{- 3} \times (- 10^{3}) = V_{B}$
$V_{B} - V_{A} = 15$