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The switch in the circuit of Fig. 30-15 has been closed on a for a very long time when it is then thrown to b. The resulting current through the inductor is indicated in Fig. 30-28 for four sets of values for the resistance R and inductance L: (1) R=R0, (2) 2R0=R, (3) R0and2L0 , (4) 2R0and2L0. Which set goes with which curve?

Short Answer

Expert verified

(1) curve a indicates set 2, (2) curve b indicates set 4, (3) curve c indicates set 1, and

(4) curve d indicates the set 3.

Step by step solution

01

Step 1: Given

  1. Fig. 30-28.
  2. Fig.30-15.
  3. The switch has been closed on ‘²¹â€™ for a very long time when it is then through to ‘²ú’.
  4. In set (1)R=R0andL=L0.
  5. In set (2)R=2R0andL=L0.
  6. In set (3)R=R0androle="math" localid="1661834742211" L=2L0.
  7. In set (4) R=2R0 and L=2L0.
02

Determining the concept

Applying Ohm’s law and using Eq.30-42, it can find the induced current i and the inductive time constant τLfor corresponding given sets. Comparing these values with the given curves in Fig.30-28, find which curve indicates which set.

Formulae are as follow:

  1. According to Ohm’s law, the current is,

i=εindR.

  1. From Eq.30-41, the current is,

i=εR1-e-t/TL

  1. The inductive time constant is given by,

τL=LR

03

(a) Determining which set goes with which curve

From Fig.30-28, the current decreases exponentially with respect to time .

According to Ohm’s law, the current is,

i=εindR...............................................................................(1)

Where, εindis induced emf.

From Eq.30-41, the current is,

i=εR1-e-t/TL...............................................................................(30-41)

Where, τLthe inductive time constant and is given by,

τL=LR...............................................................................(30-42)

In set (1) R=R0and L=L0, from Eq.1 and Eq.30-42, it gives,

i1=εindR0

and

τ1=L0R0

In set (2) R=2R0and L=L0, from Eq.1 and Eq.30-42, it gives

i2=εind2R0

and

τ2=L02R0

In set (3) R=R0and L=2L0, from Eq.1 and Eq.30-42, it gives

role="math" localid="1661836172518" i3=εindR0

and

τ3=2L0R0

In set (4) R=2R0 and L=2L0, from Eq.1 and Eq.30-42, it gives

i4=εind2R0

and

role="math" localid="1661836360441" τ4=2L02R0

Therefore,

τ4=L0R0

Therefore, comparing the above solutions it gives,

i1=i3>i2=i4

And the inductive time constant,

τ3=τ1>τ4=τ2

If the time constant has a larger value, it means decay is slower in RL circuits, and there is fast decay in RL circuit when the time constant has smaller value. Considering this in the above solutions and from Fig.30-28, the curves it gives

c>a≈d>b.

Hence, from the above results and from Fig.30-28,

(1) curve a indicates set 2, (2)curve b indicates set 4, (3) curve c indicates set 1, and (4) curve d indicates set 3.

Using Ohm’s law and Eq.30-42, find the answer to this question.

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