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

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

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Solution

(1)Step 1: Draw the diagram

Step 2: Find the total parallel resistance
Formula used: $\frac{1}{R_{P}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}$

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 $(R_{P})$

of the combination is given by,

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

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

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

$R_{P} = \frac{20}{19} Ω$

Total resistance of the combination is $\frac{20}{19} Ω .$

$B)$ Step 1: Find the Potential drop across each resistor

Formula used: $I = \frac{V}{R}$

Emf of the battery, $V = 20 V$

Current $(I_{1})$ flowing through resistor $R_{1} = 2 Ω$ is given by,

$I_{1} = \frac{V}{R_{1}} = \frac{20}{2} = 10 A$

Current $(I_{2})$ flowing through resistor $R_{2} = 4 O m e g a$ is given by,

$I_{2} = \frac{V}{R_{2}} = \frac{20}{4} = 5 A$

Current $(I_{3})$ flowing through resistor $R_{3} = 5 Ω$ is given by,

$I_{3} = \frac{V}{R_{3}} = \frac{20}{5} = 4 A$

Final Answer: Total resistance, $R = \frac{20}{19} Ω$

Current across $2 Ω$ resistor $(I_{1}) = 10 A$

Current across $4 Ω$ resistor $(I_{2}) = 5 A$

Current across $5 Ω$ resistor $(I_{3}) = 4 A$

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