(a) Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel.

Solution:

(a) There are three resistors of resistances,

R1 = 2 Ω, R2 = 4 Ω, and R3 = 5 Ω

They are connected in parallel. Hence, total resistance (R) of the combination is given by,

$\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}$

$=\frac{1}{2}+\frac{1}{4}+\frac{1}{5}=\frac{10+5+4}{20}=\frac{19}{20}$

$\therefore R=\frac{20}{19} \Omega$

Therefore, total resistance of the combination is $\frac{20}{19} \Omega$.

(b) Emf of the battery, V = 20 V

Current (I1) flowing through resistor R1 is given by,

$I_{1}=\frac{V}{R_{1}}$

$=\frac{20}{2}=10 \mathrm{~A}$

Current $\left(l_{2}\right)$ flowing through resistor $R_{2}$ is given by,

$I_{2}=\frac{V}{R_{2}}$

$=\frac{20}{4}=5 \mathrm{~A}$

Current (I3) flowing through resistor R3 is given by,

$I_{3}=\frac{V}{R_{3}}$

$=\frac{20}{5}=4 \mathrm{~A}$

Total current, I = I1 + I2 + I3 = 10 + 5 + 4 = 19 A

Therefore, the current through each resister is 10 A, 5 A, and 4 A respectively and the total current is 19 A.