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

ii) 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

Given, three resistors 2 $Ω, 4 Ω and 5 Ω$ are combined in parallel.

Therefore,
Total resistance of parallel combination,
$\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} or \frac{1}{R} = \frac{1}{2} + \frac{1}{4} + \frac{1}{5} = \frac{10 + 5 + 4}{20} = \frac{19}{20} Ω$

$R = \frac{20}{19} Ω$

ii) The combination is connected to a battery of 20 V.
Let the current through resistances 2Ω, 4Ω and 5Ω are I₁, I₂ and I₃ respectively.
Now, using Ohm's law across each resistor we get,
$I_{1} = \frac{V}{R_{1}} = \frac{20}{2} = 10 A I_{2} = \frac{V}{R_{2}} = \frac{20}{4} = 5 A I_{3} = \frac{V}{R_{3}} = \frac{20}{5} = 4 A$
Hence,
Total current is given by
$I = I_{1} + I_{2} + I_{3} = 10 + 5 + 4 = 19 A .$

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