Loop 1

14 - 2(I1 - I3) - 1(I1 - I2) = 0

therefore

3I1 - I2 -2I3 = 14[Eqn. 1]

Loop 2

1(I2 - I1) + 1(I2 - I3) + 2I2 = 0

therefore

I1 -4I2 + I3 = 0 [Eqn. 2]

Loop 3

1(I3 - I2) + 2(I3 - I1) + I3 = 0

therefore

2I1 + I2 - 4I3 = 0 [Eqn. 3]
 

Solve Eqns. 1, 2, 3 for the currents.

From [2] I1 = 4I2 - I3, substitute this in [1] and [2] and get

11I2 - 5I3 = 14 [Eqn. 4]

and

I2 = (2/3)I3 [Eqn. 5]
 

Solve [4] and [5], to get

I3 = 6 amp

I2 = 4 amp

I1 = 10 amp
 

Equivalent R = V/I where I is the total current out of the battery, i.e., I1

R = 14/10 = 1.4 W
 

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