91Ó°ÊÓ

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

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

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

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

$R_{eq}=R_{1,eq}+4\mathrm{Î\copyright }+2\mathrm{Î\copyright }=4\mathrm{Î\copyright }+4\mathrm{Î\copyright }+2\mathrm{Î\copyright }=10\mathrm{Î\copyright }$

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

$I=\frac{\mathrm{Î\mu }}{R_{eq}}=\frac{100V}{10\mathrm{Î\copyright }}=10A$

Using Ohm's law we get the voltage across $4\mathrm{Î\copyright }$resistor

$â–³V_4=I{R}_4=\left(10A\right)\left(4\mathrm{Î\copyright }\right)=40V$

Point c of the circuit is grounded, so its voltage is zero. The resistor $4\mathrm{Î\copyright }$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 $2\mathrm{Î\copyright }$resistor we use Ohm's law

$â–³V_4=I{R}_4=\left(10A\right)\left(4\mathrm{Î\copyright }\right)=40V$

The $20\mathrm{Î\copyright }$resistor is located between a and $b$, so the voltage at $b$is

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

$20\mathrm{Î\copyright }$Resistor is located between points b and c, so the voltage across is the potential difference between both points

$V_{20}=V_b-V_c=40V -0V=40V$

To get current through $20\mathrm{Î\copyright }$we use Ohm's law

$I_{20}=\frac{â–³V_{20}}{R_{20}}=\frac{40V}{20\mathrm{Î\copyright }}=2A$