Problem 21.29

21.29. b)
To start with, the $20 \Omega$ and $5 \Omega$ resistors that are on the large loop in figure 21.29 are in series. They are equivalent to one $25 \Omega$ resistor. That resistor is in parallel with the $10 \Omega$ resistor that isn't on the same wire with the battery and the other $5 \Omega$ resistor. Those three together can be thought of as in series with the $10 \Omega$ resisitor next to the battery. In equation terms
$R_{equivalent} = 10 \Omega + \frac{1}{ \left(\frac{1}{10 \Omega} + \frac{1}{5 \Omega} + \frac{1}{20 \Omega + 5 \Omega}\right) }$
$R_{equivalent} = 12.94 \Omega$
In the equivalent circuit, the current is given by: $I = \frac{V}{R}$
$I = \frac{25 V}{12.94 \Omega}$
I = 1.93 A
We can now just look at the voltage drop along the circuit that contains the $10 \Omega$ resistor next to the battery, the battery, and the equivalent resistance from the other branches of the circuit. The resisitance from the other resistors other than the $10 \Omega$ resisitor next to the battery is
$R = 2.94 \Omega$
So, we can solve for the potential drop across that resistor. We find
V = IR
$V = (1.93 A)(2.94 \Omega)$
V = 5.68 V
Since this is the same voltage drop across all the all three branches of the circuit that are in parallel, then the voltage drop across the points a to b is given by
V_ab = 5.68 V

b)
Since the current through the $20 \Omega$ resistor is the same as that along the segment ab, we find that
$I = \frac{V_{ab}}{R_{ab}}$
$I_{20} = \frac{5.68 V}{25 \Omega}$
I₂₀ = .227 A