Determine the current through the resistor R2 in the circuit of Fig. 7.12.

Question

Determine the current through the resistor [latex] R_2 [/latex] in the circuit of Fig. 7.12.

Accepted Answer

There are two meshes whose currents are [latex]I_{1} and I_{2}[/latex], respectively, and we note that the mesh current [latex]I_{1}[/latex] is equal to the current source [latex]I_{s}[/latex]. Applying KVL to mesh 2 and taking the anticlockwise direction to be positive, we have

[latex]V_{s} + R_{3}I_{2} + R_{2}I_{3} = 0[/latex]

Since [latex]I_{3} = I_{2} - I_{1}[/latex] then

[latex]V_{s} +R_{3}I_{2} + R_{2}I_{2}-R_{2}I_{1} =0[/latex]

[latex](R_{2}+R_{3})I_{2}=R_{2}I_{1}-V_{s} [/latex]

[latex]= R_{2} I_{s} - V_{s}[/latex]

[latex]I_{2} = ( R_{2} I_{s} - V_{s})/(R_{2} + R_{3})[/latex]

Putting in the values we have

[latex]I_{2} = [(2 ×1) - 6)]/(2 + 1) = ( - 4 / 3 ) A[/latex]

The current through the resistor [latex]R_{2}[/latex] is

[latex]I_{3} = I_{2} - I_{1} = I_{2} - I_{s} =( - 4 / 3 ) - 1 =-7/3 = -2.33v A[/latex]

The negative sign indicates that the current is flowing in the opposite direction to that shown in the circuit diagram (i.e. it flows downwards through the
resistor).

Question 7.15: Determine the current through the resistor R2 in the circuit...

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