# Find current in the circuit using mesh analysis?

### 4 Answers

- RogerLv 74 weeks ago
i1 = -1.75 amps and i2 = 1.25 amps

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let V1 be the voltage across the 3 amp current source

first loop equation

5 volts = 5(i1) + V1

second loop equation

V1 = 3(i2) + 10 volts

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now substituting for V1 in the first loop equation

5 volts = 5(i1) + 3(i2) + 10 volts

- 5 volts = 5(i1) + 3(i2)

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3 amps = i2 - i1

i2 = 3 amps + i1

-5 volts = 5(i1) + 3( 3 +i1)

-14 volts = 8(i1)

i1 = -14/8 amps = -1.75 amps

i2 = 3 +(-1.75 amps) = 1.25 amps

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to check find V1

V1 = 10 volts + 3(i2) = 13.75 volts

V1 also equals

V1 = 5 volts -5 (i1) = 5 volts -(-1.75)5 = 13.75 volts so our answer is right

- oldschoolLv 74 weeks ago
This one is much easier using node analysis. Let bottom of the ckt = ground and just above the 3A source = node V. Then use KCL to sum the currents leaving

node V to = 0:(V-5)/5Ω - 3 + (V-10)/3Ω = 0 Multiply by 15

3V-15 - 45 +5V -50 = 0

8V = 110 so V = 110/8

The current leaving node V left into the 5Ω = (13.75-5)/5 = 1.75A

The current leaving node V right into the 3Ω = (13.75-10)/3 = 1.25A

Compare the power delivered by the sources to the power consumed by resistance:

-5*1.75 -10*1.25 + 13.75*3 = 20W

Notice both voltage sources are being "charged" and only the current source is delivering power. Now find the power consumed by resistance using P = i²*R:

1.75²*5Ω + 1.25²*3Ω = 20W Checks

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- AshLv 74 weeks ago
From middle branch we get i₂ - i₁ = 3 A

i₂ = 3 + i₁ ...(1)

Kirchhoff's Voltage Law for outer mesh

5 - 5i₁ - 3i₂ - 10 = 0 ...

- 5i₁ - 3i₂ - 5 = 0

5i₁ + 3i₂ = - 5

Plug in (1)

5i₁ + 3(3 + i₁) = - 5

5i₁ + 9 + 3i₁ = - 5

8i₁ = -14

i₁ = -14/8

i₁ = - 1.75 A

i₂ = 3 + (-1.75) = 1.25 A

So current i₁ is 1.75 A flowing in opposite direction as shown in mesh 1

while current i₂ is 1.25 A flowing in same direction as shown in mesh 2

Glad you note the incorrect current flow in one loop. That is first thing I would address--or the signs are backwards.