Chapter 30: Q95P (page 902)

In Fig. 30-77, $R_{1} = 8.0 Ω, R_{2} = 10 Ω, L_{1} = 0.30 H, L_{2} = 0.20 H$ and the ideal battery has $ε = 6.0 V$ . (a) Just after switch S is closed, at what rate is the current in inductor 1 changing? (b) When the circuit is in the steady state, what is the current in inductor 1?

Short Answer

Expert verified

a) The rate of change of current in inductor 1 is,

$\frac{d i_{b a t}}{d t} = 20 A / s$

b) The current in the inductor 1 is $i = 0.75 A$

Step by step solution

Given

$R_{1} = 8.0 Ω R_{2} = 10 Ω L_{2} = 0.20 H ε = 6.0 V$

Understanding the concept

By using the equation 30-35, we can find the rate of current. After that by using the KVL

We can find the current in the inductor 1.

Formula:

$i_{b a t} = \frac{ε}{R} (1 - e^{\frac{R t}{L}}) \frac{d i_{b a t}}{d t} = \frac{|ε|}{L_{1}}$

(a) Calculate the rate of change of current in inductor 1.

We have, equation for self-induced emf as,

$L_{1} \frac{d i_{b a t}}{d t} = |ε|$

So that the rate of current is,

$\frac{d i_{b a t}}{d t} = \frac{|ε|}{L_{1}}$

By substituting the value

$\frac{d i_{b a t}}{d t} = \frac{6.0}{0.30} \frac{d i_{b a t}}{d t} = 20 A / s$

(b) Calculate the current in the inductor 1 at steady state.

Now we have to find the current in inductor 1 so that by Appling the KVL for outer loop we can writer as

$ε - {iR}_{1} - |ε_{L_{1}}| - |ε_{L_{2}}| = 0$

But as $t \to \infty e m f$ in the coil vanishes

$ε - i R_{1} = 0 ε - {iR}_{1} i = \frac{ε}{R_{1}} i = \frac{6.0}{8.0} i = 0.75 A$

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In Fig. 30-77, $R_{1} = 8.0 Ω, R_{2} = 10 Ω, L_{1} = 0.30 H, L_{2} = 0.20 H$ and the ideal battery has $ε = 6.0 V$ . (a) Just after switch S is closed, at what rate is the current in inductor 1 changing? (b) When the circuit is in the steady state, what is the current in inductor 1?

Short Answer

Step by step solution

Given

Understanding the concept

(a) Calculate the rate of change of current in inductor 1.

(b) Calculate the current in the inductor 1 at steady state.

One App. One Place for Learning.

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In Fig. 30-77,R1=8.0Ω,R2=10Ω,L1=0.30H,L2=0.20Hand the ideal battery hasε=6.0V. (a) Just after switch S is closed, at what rate is the current in inductor 1 changing? (b) When the circuit is in the steady state, what is the current in inductor 1?

Short Answer

Step by step solution

Given

Understanding the concept

(a) Calculate the rate of change of current in inductor 1.

(b) Calculate the current in the inductor 1 at steady state.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Mechanics and Materials

Electricity and Magnetism

Circular Motion and Gravitation

Waves Physics

Oscillations

Electrostatics

Study anywhere. Anytime. Across all devices.

In Fig. 30-77, $R_{1} = 8.0 Ω, R_{2} = 10 Ω, L_{1} = 0.30 H, L_{2} = 0.20 H$ and the ideal battery has $ε = 6.0 V$ . (a) Just after switch S is closed, at what rate is the current in inductor 1 changing? (b) When the circuit is in the steady state, what is the current in inductor 1?