In the figure R₁ = 9.69 kQ2, R₂ = 15.9 kQ, C = 0.379 μF, and the ideal battery has emf ε =20.0 V. First, the switch is closed a long time so that the steady state is reached. Then the switch is opened at time t = 0. What is the current in resistor 2 at t = 3.90 ms?

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In the figure R₁ = 9.69 kN, R₂ = 15.9 kn, C = 0.379 μF,
and the ideal battery has emf ε =20.0 V. First, the
switch is closed a long time so that the steady state is
reached. Then the switch is opened at time t = 0.
What is the current in resistor 2 at t = 3.90 ms?
E
Number i
+
Fu
R₁
www
R₂
с
Units A
Transcribed Image Text:In the figure R₁ = 9.69 kN, R₂ = 15.9 kn, C = 0.379 μF, and the ideal battery has emf ε =20.0 V. First, the switch is closed a long time so that the steady state is reached. Then the switch is opened at time t = 0. What is the current in resistor 2 at t = 3.90 ms? E Number i + Fu R₁ www R₂ с Units A
Expert Solution
Step 1: INTRODUCTION

Given Data:

The resistance of resistor 1 is R subscript 1 equals 9.69 space k capital omega equals 9.69 cross times 10 cubed space capital omega

The resistance of resistor 2 is R subscript 2 equals 15.9 space k capital omega equals 15.9 cross times 10 cubed space capital omega

The capacitance of the capacitor is C equals 0.379 space mu F equals 0.379 cross times 10 to the power of negative 6 end exponent space F

The emf of the battery is epsilon equals 20 space V

The time is t equals 3.9 space m s equals 3.9 cross times 10 to the power of negative 3 end exponent space s

There will be no current in the capacitor branch because the switch is opened.

So at this condition the voltage across capacitor is same as the vlotage across resistor 2.



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