Below is a circuit schematic of sources and resistors.

$$V_S=10V, R_1=50Ω, R_2=15Ω, R_3=5Ω, I_S=2A$$

enter image description here

Question 1: Calculate the current i1 in amps that goes through R3 from left to right.

Question 2: For the same circuit, calculate the voltage across R3.

I used mesh analysis to solve this problem. When I assign the current in the counter clock wise (CCW) direction, I get the correct answer. But if I assign i1 in the clockwise (CW) direction, I cannot get the same answer. It appears I am making an error when I use Ohm's law across R2.

When i1 is CCW, \$i_1\$ and \$i_s\$ are in opposite directions and the voltage is \$(i_1-i_s)R_2\$. When i1 is CW, \$i_1\$ and \$i_s\$ are both in the same direction and pointing downwards. So I thought the voltage would be \$(i_1+i_s)R_2\$.

But when I do this the answer is incorrect. Can someone please explain why? I think I should get the same answer no matter which way I assign the current.

enter image description here


1 Answer 1


The equation for clockwise should be:

$$-v_{s} + (i_1 + i_s)R_2 + i_1R_3 = 0$$ $$-v_s + i_1(R_2+R_3) + i_sR_2 = 0$$ $$i_1 = - \frac{ i_sR_2 - v_s}{R_2+R_3}$$

Which is exactly negative of the other direction choice.

  • \$\begingroup\$ thank you. What about the fact that the polarity of the R3 was explicit on the diagram? This can be ignored? Your equation does not recognize this. \$\endgroup\$ Commented Jan 19, 2015 at 23:05
  • 2
    \$\begingroup\$ That was likely put on there to trick you, which it did. Just think logically: if R2 is added with a positive I1, why would R3 be subtracted with a negative I1? The polarity of V3 is arbitrary, it is the direction of the current through the resistor that you should pay attention to (or in this case, the direction you assumed the current to be flowing for your calculations). Resistors are not voltage sources. It was added to make sure you actually understand the process you're using, instead of just copying from an equivalent problem. \$\endgroup\$
    – I. Wolfe
    Commented Jan 19, 2015 at 23:23

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