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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.

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    \$\begingroup\$ How did you get the first loop equation? \$\endgroup\$
    – Chu
    Oct 26 '21 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
    Oct 26 '21 at 18:24
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\$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

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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

V(source)=-V2+V3

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

6=-8i2+4i3

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

V(Source2)+V2-V3=0.

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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