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

Byju's Answer

Standard XII

Physics

Inductors in Circuits

In L - R circ...

Question

In $L - R$ circuit shown in diagram, reading of ammeter is $i_{1}$ . Now, a diamagnetic material is inserted into the inductor, then the new reading of ammeter is $i_{2},$ will be

i_{1} > i_{2}

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

i_{2} > i_{1}

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

i_{1} = i_{2}

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

cannot say about

i_{1} a n d i_{2},

it depends upon value of R

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

Open in App

Solution

The correct option is B $i_{2} > i_{1}$
If diamagnetic material is introduced,
$L ↓ \Rightarrow Z ↓ \Rightarrow i ↑$

Suggest Corrections

Similar questions

Q. In

L - R

circuit shown in diagram, reading of ammeter is

i_{1}

. Now, a diamagnetic material is inserted into the inductor, then the new reading of ammeter is

i_{2},

will be

Q. In the transistor configuration,

I_{1}, I_{2}

and

I_{3}

are ammeter readings. Then

Q. A student using the same two resistors, ammeter, voltmeter and battery, sets up two circuits A and B connecting the two resistors first in series and then in parallel.

If the ammeter and voltmeter readings in the two cases be I₁, I₂ and V₁, V₂ respectively he is likely to observe that :
(1) I₁ = I₂ but V₁ ≠ V₂
(2) I₁ < I₂ but V₁ = V₂
(3) I₁ > I₂ but V₁ = V₂
(4) I₁ = I₂ and V₁ = V₂

Let $I_{1} = {(\frac{π}{4})}^{2} + \sqrt{2}, I_{2} = {(t a n^{- 1} (\frac{1}{e}))}^{2} + \frac{2 e}{\sqrt{e^{2} + 1}}, I_{3} = (t a n^{- 1} e)^{2} + \frac{2}{\sqrt{e^{2} + 1}}$ , then which of the following is true?

Q. For three circuits, shown here

the same two resistors R₁ and R₂ have been connected in parallel in all the circuits but the voltmeter and the ammeter have been connected in three different positions. The relation between the three voltmeter and ammeter readings would be :
(1) V₁ = V₂ = V₃ and I₁ = I₂ = I₃
(2) V₁ ≠ V₂ ≠ V₃ and I₁ = I₂ = I₃
(3) V₁ = V₂ = V₃ and I₁ ≠ I₂ ≠ I₃
(4) V₁ ≠ V₂ ≠ V₃ and I₁ ≠ I₂ ≠ I₃

Join BYJU'S Learning Program

In L−R circuit shown in diagram, reading of ammeter is i1. Now, a diamagnetic material is inserted into the inductor, then the new reading of ammeter is i2, will be

The correct option is B i2>i1If diamagnetic material is introduced, L ↓ ⇒ Z ↓ ⇒ i ↑

In $L - R$ circuit shown in diagram, reading of ammeter is $i_{1}$ . Now, a diamagnetic material is inserted into the inductor, then the new reading of ammeter is $i_{2},$ will be

The correct option is B $i_{2} > i_{1}$
If diamagnetic material is introduced,
$L ↓ \Rightarrow Z ↓ \Rightarrow i ↑$