Question

Find current in the 4 ohm resistor by any three methods.

Answer 1

Writing down circuit equations, with given conditions, and marking three clockwise loop-currents as i₁, i₂ and i₃. 

i₁ = 5A, due to the current source of 5 Amp 

V_A −V_B = 6V, due to the voltage source of 6Volts 

i₃ −i₂ = 2A, due to the current source of 2 Amp. 

V_A = (i₁ −i₂)2, V_B = i₃ × 4 

With these equations, the unknowns can be evaluated. 

2 (i₁ −i₂) − 4 i₃ = 6, 2 (5 −i₂) − 4(2 + i₂) = 6 

This gives the following values : i₂ = − 2/3 Amp., i₃ = 4/3 Amp. 

V_A = 34/3 volts, V_B = 16/3 volts 

Method 2 : Thevenin’s theorem : Redraw the circuit with modifications as in Fig. 

R_TH = + 14 − 6 = 8 V 

R_TH = 2 ohms, looking into the circuit form X-Y terminals after de-activating the sources 

I_L = 8/(2 + 4) = 4/3 Amp. 

Method 3 : Norton’s Theorem : Redraw modifying as in Fig.

IN = 2 + 2 = 4 Amp. This is because, X and Y are at ground potential, 2-ohm resistor has to carry 3 A and hence from 5-Amp. source, 2-Amp current is driven into X-Y nodes. 

R_N = 2 ohms Then the required current is calculated as shown in Fig.