Problem 21.37

21.37. a)
Let i₁ go from the $2 k\Omega$ resistor to the 70 V battery. Let i₂ go from the 60 V battery to the $3 k\Omega$ resistor. Let i₃ go from the 80 V battery to the $4 k\Omega$ resistor. Then, we have
i₂ = i₁ + i₃
And from Kirchhoff's laws,
$70 - 60 - i_{2}(3 k\Omega) - i_{1}(2 k\Omega) = 0$
$80 - i_{3}(4 k\Omega) - 60 - i_{2}(3 k\Omega) = 0$
Solving the system of three equation and three unknowns, we find that
i₁ = .385 mA through R₁
i₂ = 3.08 mA through R₂
i₃ = 2.69 mA through R₃

b)
The potential drop from c to f is given by
$V_{cf} = -60 V - (3.08 mA)(3 k\Omega) = -69.2 V$
Therefore we conclude that point c is at a higher potential than point f.