- Three resistors $2\ \Omega,\ 4\ \Omega\ \text{and}\ 5\ \Omega$ are combined in parallel. What is the total resistance of the combination?
- 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.
Exercise
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- Given, three resistors $2\ \Omega,\ 4\ \Omega\ \text{and}\ 5\ \Omega$ are combined in parallel.
Therefore,
Total resistance of parallel combination,
$\frac{1}{\text{R}}=\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}+\frac{1}{\text{R}_3}$
$\text{or}\ \ \frac{1}{\text{R}}=\frac{1}{2}+\frac{1}{4}+\frac{1}{5}$
$=\frac{10+5+4}{20}$
$=\frac{19}{20}\ \Omega$
$\text{R}=\frac{20}{19}\ \Omega$
- The combination is connected to a battery of 20 V.
Now, using Ohm's law across each resistor we get,
$\text{I}_1=\frac{\text{V}}{\text{R}_1}=\frac{20}{2}=10\ \text{A}$
$\text{I}_2=\frac{\text{V}}{\text{R}_2}=\frac{20}{4}=5\ \text{A}$
$\text{I}_3=\frac{\text{V}}{\text{R}_3}=\frac{20}{5}=4\ \text{A}$
Hence,
Total current is given by
$\text{I}=\text{I}_1+\text{I}_2+\text{I}_3$
$=10+5+4$
$=19\ \text{A}.$
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