What is the current through the 20 Ω resistor in FIGURE P28.61?

We need to find the current through the 20 Ω resistor.

To find the total current, Both two resistors $5\Omega and20\Omega$are in parallel,so we need to find their equivalent resistance.

$$\begin{gathered} \frac{1}{R_{eq}}=\frac{1}{5\Omega }+\frac{1}{20\Omega } \\ {R}_{eq}={\left(\frac{1}{5\Omega }+\frac{1}{20\Omega }\right)}^{-1} \\ {R}_{1,eq}=4\Omega \end{gathered}$$

$4\Omega and2\Omega$resisters are both connected in series with $R_{1,eq}$, So the total resistance in the circuit is

$R_{eq}=R_{1,eq}+4\Omega +2\Omega =4\Omega +4\Omega +2\Omega =10\Omega$

The current is the same for resistors in series. so, The circuit current is calculated by using Ohm's law

$I=\frac{\epsilon }{R_{eq}}=\frac{100V}{10\Omega }=10A$

Using Ohm's law we get the voltage across $4\Omega$resistor

$△V_4=I{R}_4=\left(10A\right)\left(4\Omega \right)=40V$

Point c of the circuit is grounded, so its voltage is zero. The resistor $4\Omega$is located between the two points c and d, so the voltage at point d is

$V_4=40V$

In this case, the battery of $100V$is located between the two points a and d. At point d the voltage is $40V$, so the potential at point a is

$V_a=100V-V_d=100V-40V=60V$

To find voltage across $2\Omega$resistor we use Ohm's law

$△V_4=I{R}_4=\left(10A\right)\left(4\Omega \right)=40V$

The $20\Omega$resistor is located between a and $b$, so the voltage at $b$is

$V_b=V_a-20V=60V-20V=40V$

$20\Omega$Resistor is located between points b and c, so the voltage across is the potential difference between both points

$V_{20}=V_b-V_c=40V\hspace{0.17em}-0V=40V$

To get current through $20\Omega$we use Ohm's law

$I_{20}=\frac{△V_{20}}{R_{20}}=\frac{40V}{20\Omega }=2A$