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Three resistors \(\text{2 $\Omega $, 4 $\Omega $}\)  and \(\text{5 $\Omega $}\)  are combined in parallel.
If the combination is connected to a battery of emf \(20\text V\)  and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.

Answer

Here we  are given three Resistors \(\text{(R1, R2}\)  and \(\text{R3)}\)  as \(2,4\)  and \(5\)  ohms.
\(\mathit{R1}=2\mathit{ohm}\)
\(\mathit{R2}=4\mathit{ohm}\)
\(\mathit{R3}=5\mathit{ohm}\)
When the resistors are in parallel, the resistance is added reversely.
\(R=\frac 1{\mathit{R1}}+\frac 1{\mathit{R2}}+\frac 1{\mathit{R3}}\)
\(R=\frac12+\frac 1 4+\frac 1 5\)
\(R=\frac{20}{19}\mathit{ohms}\)
Hence in series, the resistance will be \(\frac{20}{19}\mathit{ohms}\) .
The total current in the resistor may be calculated as-
\(V=I R\)
\(20=\frac{I 20}{19}\)
\(I=19A\)
To calculate current in each resistor,
Resistor \(1\)
\(\mathit{I1}=\frac V{\mathit{R1}}\)
\(\mathit{I1}=\frac{20} 2=10A\)
Resistor 2
\(\mathit{I2}=\frac V{\mathit{R2}}\)
\(\mathit{I2}=\frac{20} 4=5A\)
Resistor 3
\(\mathit{I3}=\frac V{\mathit{R3}}\)
\(\mathit{I3}=\frac{20} 5=4A\)
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