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Solve using KVL and KCL and find voltages and currents.

For this I took $$i_1=i_2+i_3$$ $$v_1=2i_1$$ $$v_2=8i_2$$ $$v_3=4i_3$$

The equations I got by assuming loop 1 and loop 2 were $$5i_2 + i_3 = 5$$ and $$-2i_3 + 4i_2 = -3$$

On solving I'm getting \$i_2=0.5\text{ A}\$ and \$i_3=2.5\text{ A}\$ but the answer key says \$i_2=500\text{ mA}\$ and \$i_3=1.25\text{ A}\$.

I don't understand where my mistake is.

  • 1
    \$\begingroup\$ How did you get the first loop equation? \$\endgroup\$
    – Chu
    Commented Oct 26, 2021 at 7:38
  • \$\begingroup\$ I'll just tell you that if you treat the bottom wire as ground (0 V) then the node shared by all three resistors is at +4 V (from simple 3-resistor divider equation.) From there you can compute the indicated currents. I think i3 is 2.5 A, as well. \$\endgroup\$
    – jonk
    Commented Oct 26, 2021 at 18:24

2 Answers 2


\$i_1\$, \$i_2\$ and \$i_3\$ are branch currents, not loop currents. Let's write equations.

eq1 = -10 + 2 i1 + 8 i2 == 0 (KVL loop 1)

eq2 = 4 i3 - 6 - 8 i2 == 0 (KVL loop 2)

Since there are three unknowns, there must be three equations.

eq3 = i1 == i2 + i3 (KCL at top node)

sol = Solve[{eq1, eq2, eq3}, {i1, i2, i3}]

{{i1 -> 3, i2 -> 1/2, i3 -> 5/2}}

Check: {eq1, eq2, eq3} /. sol

{{True, True, True}}

You are right, the answer key is wrong: CONGRATS


I don't understand where you got your equations from TBH.... Maybe if I pulled out a pencil and paper

But I'd do this:

Your first 3 eq are correct.

So loop 1 is:

V(source1)= V1+V2

10=2i1+ 8i2

Note the DIRECTION of the voltages in each resistor. They both oppose the source so they're both positive

Loop 2 is


The "-" before the V2 is critical. This is written because that voltage (as defined) ADDS to the voltage source


If it's clearer, look at it as Kirchoffs "The voltages must equal zero"


V3 in that form opposes the loop, so it's negative.

There's a 3rd loop that runs around the entire perimeter. You can write an eqation for that too...

V(source1)=V2+V3-V(source2).... again the (-) is key here, because of the direction of the voltages defined

So you now have 3 equations with 3 unknowns.

Does this help?

If not, post more of your work. Just posting the final equations doesn't help find the problem....


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