Problem 5.3-2

For the current source, use a source transformation to get a voltage source, with the following characteristics:

V_s = i_s*R_p = 8*2 = 16 V and R_s = R_p = 8Ω

We then have in series two resistors, one of 8Ω and the other of 4Ω, which gives R₁ = 12Ω

We then have the first circuit:

We can then use a source transformation again for the right hand side one, and we get the next circuit, where the current i_s is such that

i_s = v_s/R_s = 16/12 = 4/3 A

The two parallel resistors (6Ω and 12Ω) give one equivalent resistor of R_eq such that 1/R_eq = 1/6 + 1/12 = 3/12 = 1/4 so R_eq = 4Ω

Now we have the following circuit:

Applying the KVL in the left loop we have:

10 – 3*ia – 4*(ia – i_s) = 0

thus 10 = 7*ia –4* i_s = 7*ia – 16/3

 

so we get ia = 46/21 A