**Question:**

(a) Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel. What is the total resistance of the combination?

(b) If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.

**Solution:**

(a) There are three resistors of resistances,

*R*1 = 2 Ω, *R*2 = 4 Ω, and *R*3 = 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 (*I*1) flowing through resistor *R*1 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 (*I*3) flowing through resistor *R*3 is given by,

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

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

Total current, *I* = *I*1 + *I*2 + *I*3 = 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.

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