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

Byju's Answer

Standard XII

Physics

Time Constant

The circuit s...

Question

The circuit shown in the figure is a part of a big network. If at a certain instant of time, the current $(i)$ through it is $5 A$ and is decreasing at the rate of $10^{3} A/s$ , then $V_{B} - V_{A}$ at that instant is -

50 V

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

10 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

20 V

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

Open in App

Solution

The correct option is C $15 V$

According to KVL,

$V_{A} - i R + E - L \frac{d I}{d t} = V_{B}$

$\Rightarrow V_{A} - 5 \times 1 + 15 - 5 \times 10^{- 3} \times (- 10^{3}) = V_{B}$

$∴ V_{B} - V_{A} = 15 V$

Hence, option $(C)$ is the correct answer.

Suggest Corrections

Similar questions

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 5A and it is decreasing at a rate of

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

then

V_{B} - V_{A}

equals

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 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

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

Join BYJU'S Learning Program

The circuit shown in the figure is a part of a big network. If at a certain instant of time, the current (i) through it is 5 A and is decreasing at the rate of 103 A/s, then VB−VA at that instant is -

The correct option is C 15 V According to KVL, VA−iR+E−LdIdt=VB ⇒VA−5×1+15−5×10−3×(−103)=VB ∴VB−VA=15 V Hence, option (C) is the correct answer.

The circuit shown in the figure is a part of a big network. If at a certain instant of time, the current $(i)$ through it is $5 A$ and is decreasing at the rate of $10^{3} A/s$ , then $V_{B} - V_{A}$ at that instant is -

The correct option is C $15 V$

According to KVL,

$V_{A} - i R + E - L \frac{d I}{d t} = V_{B}$

$\Rightarrow V_{A} - 5 \times 1 + 15 - 5 \times 10^{- 3} \times (- 10^{3}) = V_{B}$

$∴ V_{B} - V_{A} = 15 V$

Hence, option $(C)$ is the correct answer.