# solve by super node analysis?

10V 5.6k 1k 5.6k 5v 1k Relevance

Label the three nodes

V1 ( the node at the positive end of battery V1)

V2 ( the node at the positive end of battery V2)

V3 ( the node at the negative end of battery V2)

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Node V1 Note V1 is 10 volts

Iv1 = (10 V)/ (5.6 K) + (10 - V2)/(1 K) where Iv1 is current into node V1

Node V2

(10- V2)/(1 K) = V2/(5.6 K) + iV2 where IV2 is the current into the V2 source

Node V3

IV2 = V3/(1 K)

note V3 = V2 - 5 Volts

So IV2 = (V2 -5)/(1 K)

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combine the V2 and V3 node equations

(10 - V2)/(1 K) = V2/ (5.6 K) + (V2 - 5)/ (1 K)

10/(1 K) + 5/(1 K) = 2(V2)/(1 K) + V2/(5.6 K)

multiply by 1,000 to remove the K

15 = (2.179)V2

V2 = 6.89 Volts

round off to

V2 = 6.9 Volts

V3 = 1.9 Volts

V1 = 10 Volts

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I R1 = 10/(5.6 K) = 1.8 mA

IV2 = V3/(1K) = 1.9 mA this is I R4

I R2 = 6.9 V/(5.6K) = 1.2 mA

I R3 = (10v- 6.9v) /(1k) = 3.1 mA flowing to the right

and

IV1 = I R2 + I R3 = 4.9 mA

• Let the node left of the 5v source be V.

We know the left side of the ckt =10v. Sum the currents leaving V to = 0

(V-10)/1kΩ + V/5.6Ω + (V-5)/1kΩ = 0 Multiply by 56

56V-560+10V+56V-280 = 0

122V = 840 -----> V = 840/122 = 420/61 or about 6.9V

The current out of the 10v source = 10/5.6k+(10-6.9)/1k = 4.9mA

The current down into the left 5.6k = 10/5.6k = 25/14 =1.79mA

The current left to right in the 1kΩ = (10-420/61)/1kΩ = 3.11mA

The current down into the next 5.6k right = (420/61)/5.6k = 1.23mA

The current right into the 1k = (420/61)/1k = 1.89mA

The voltage across the  top 1k = 190/61 = 3.11v

The voltage across the right 1k = 1.89v