Question in the attachment Diagram 1R1AMAR3R2 Refer to diagram 1.In the circuit shown: R1 = 1.45 Q, R2 = 7.04 Q, R3 = 6.3 Q, L = 20 mH and V= 37.7 V. Initially switch S is open; at t = 0 the switch is closed. At that instant, find 12, thecurrent through R2 in A.NOTE: Capacitor 18 µF has been removed from the problem.

Answered: Refer to diagram 1. In the circuit…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Question in the attachment

Refer to diagram 1.
In the circuit shown: R1 = 1.45 Q, R2 = 7.04 Q, R3 = 6.3 Q, L = 20 mH and V= 37.7 V. Initially switch S is open; at t = 0 the switch is closed. At that instant, find 12, the
current through R2 in A.
NOTE: Capacitor 18 µF has been removed from the problem.

Expert Solution

Step 1

Kirchhoff's loop rule states that in any closed path of an electrical circuit, the algebraic sum of all the potential differences is zero, i.e., $\sum ∆ V = 0$

Step 2

At the instant the switch is closed, the inductor is against the change of current (Lenz's law)

$\Rightarrow I_{3} = 0$ ( current through R₃)

Now, since $I_{3} = 0$ , so $I_{1} = I_{2}$ . Therefore from Kirchhoff's loop law :

$\Rightarrow V - I_{1} R_{1} - I_{2} R_{2} = 0 \Rightarrow 37.7 V - I_{2} (R_{1} + R_{2}) = 0 \Rightarrow I_{2} = \frac{37.7 V}{(R_{1} + R_{2})} \Rightarrow I_{2} = \frac{37.7 V}{(1.45 Ω + 7.04 Ω)} \Rightarrow I_{2} = 4.44 A$

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