Question

Two batteries and four resistors are connected as pictured below. Clearly label three unique currents within this circuit. So
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Answer #1

Given:

R_1=2\ \Omega\\ R_2=3\ \Omega\\ R_3=6\ \Omega\\ R_4=5\ \Omega\\ V_1=4\ \text V\\ V_2=8\ \text V

Let us assume that I_1 flows to up through V_1 , I_2 flows up through RE and I_3 flows down through V_2

Applying Kirchchoff's Voltage Rule to left mesh in CW direction, we get

V_1-I_1R_1+I_2R_2-I_1R_4=0\\ I_1(R_1+R_4)-I_2R_2=V_1\\ 7I_1-3I_2=4.........(1)

Applying KVL to right mesh in CW direction, we get

-I_2R_2+V_2-I_3R_3=0\\ I_2R_2+I_3R_3=V_2\\ 3I_2+6I_3=8.........(2)

Applying Kirchchoff's Current Law to any of the junctions, we get

I_1+I_2-I_3=0.........(3)

Solving these three linear equations, we get

I_1=0.741\ \text A\\ I_2=0.395\ \text A\\ I_3=1.14\ \text A

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