In the circuit shown below both Switch A and Switch B are in

In the circuit shown below, both Switch A and Switch B are initially open, and the capacitor is initially charged to Q_0 = 36 mu C. Then, Switch A and Switch B are closed simultaneously True or False: The current I_1 through the R_1 resistor is constant as a function of time after the two switches are closed. Explain your answer briefly. Immediately after Switch A and Switch B are closed, the magnitude of the current passing through the battery is: |I_b| = 5.0 A |I_b| = 3.0 A |I_b| = 2.0 A |I_b| = 1.5 A |I_b| = 1.0A

Solution

Ans)

a) True

When both the switches are closed voltage V is directly connected in parallel with resistor R₁,Hence the following explanation follows

Because the current I₁=V/R₁ (ohm\'s law) as both V and R₁ are constant ,current is constant as a function of time

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b)

so answer is E |I_b|=1 A

Explanation

Initial voltage across capacitor is Vc=Qo/C=36u/2u=18 V

Vc=18 V

I₁=V/R₁=6/3=2A

using KVL voltage across R₂=V_R2=V-Vc=6-18=-12 V

V_R2=-12 V

I_R2=V_R2/R₂=-12/4=-3 A

Now using KCL

I_b=I_R2+I₁=-3+2=-1 A

I_b=-1 A

|I_b|=1 A

so answer is E