A linear circuit is shown in the figure.

The elements in this circuit have the following values: R = 100 Ohms, R2 = 200 Ohms, R3 = 350 Ohms, V = 5V, and I = 0.004 A.

  1. Determine the potentials of drops v_a and v_b across resistors R1 and R2 respectively.

  2. Now let us determine if the answers you came up with satisfy the laws of physics. (a) What is the power (in Watts) dissipated in resistor R1? (b) What is the power (in Watts) dissipated in resistor R2? (c) What is the power (in Watts) dissipated in resistor R3? (d) What is the power (in Watts) coming out of the voltage source V? (e) What is the power (in Watts) coming out of the current source I?

enter image description here

From the circuit I have:

\$-v_b + v_a + V = 0\$

\$v_b - V = v_a\$

\$i_1 = (v_b - V)/R_1\$

\$I + i_2 = i_1\$

\$(v_b - V)/R_1 = I + v_b/R_2\$

From this last equation I get \$v_b = 10.8\$ and hence \$v_a = 5.8\$.

However, apparently that is wrong. (And hence my answers to #2 were all wrong as well.) Why is that so? What might I be doing wrong?

  • \$\begingroup\$ Assuming \$R=R_1\$, there is only one unknown node voltage, which is \$V_\text{B}\$. (Just ground the bottom node to make it \$0\:\text{V}\$.) The result should be \$V_\text{B}=3.6\:\text{V}\$. From there, the answers just flow out. Do you see how to develop that voltage? \$\endgroup\$
    – jonk
    Commented Jan 13, 2020 at 7:54
  • \$\begingroup\$ Wrong sign in the last equation. And you haven't stated where i2 is. \$\endgroup\$
    – Chu
    Commented Jan 13, 2020 at 12:50

2 Answers 2


Here's the fully annotated circuit, which you should probably have started with yourself, since that would have made it easier to spot the error you made:


simulate this circuit – Schematic created using CircuitLab

Applying KCL to node X:

$$ I - I_1 - I_2 = 0 $$

or, as you arranged it:

$$ I - I_2 = I_1 $$

My sign for \$I_2\$ differs from yours, because your direction of \$I_2\$ is inconsistent with the polarity of voltage across \$R_2\$. Current always enters a resistor at the terminal with the higher potential, and emerges at the lower potential end.

As you can see in my schematic, my chosen direction of \$I_2\$ is consistent with this principle, flowing downwards through \$R_2\$ from its higher potential end to lower, according to my "+" and "−" annotations for \$V_B\$. Your own application of KCL implies \$I_2\$ flowing upwards, which would have it entering \$R_2\$ at the terminal marked as having the lower potential.

I could have marked the polarity of voltage across \$R_2\$ differently, with "+" at the bottom, and annotated the schematic with \$I_2\$ flowing upwards. The polarity of \$V_A\$ has changed, so the KVL equations would be different, but in the end the solution would still be correct, as long as we conform to the principle that current flows inside resistors from higher potential to lower.

Often we don't know what direction current will flow along some path, and we are forced to guess the direction. If you guessed wrong, it doesn't matter, since you'll just end up with a negative current, which is equivalent to the same positive amount flowing the other way. The arithmetic still works out.

This can be seen with \$I_1\$; in the solution \$I_1 = -14mA\$, which just tells me that my guess that current travels to the right through \$R_1\$ was incorrect. This negative sign is just telling me that actually +14mA is flowing to the left through \$R_1\$.

As a consequence of incorrectly guessing the direction of \$I_1\$, I should expect also to have incorrectly guessed the polarity of \$V_A\$. Indeed, the solution reveals \$V_A=-1.4V\$. That negative sign implies that, actually, the side of \$R_1\$ marked "+" has the lower potential.


I believe your problem is with the 4th equation. As defined in the circuit, all of the currents are entering the node so their sum must be zero.

  • \$\begingroup\$ But the currents are not defined in the circuit. There is no marking of \$i_1\$ or \$i_2\$ in the circuit. \$\endgroup\$ Commented Jan 13, 2020 at 18:31

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