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# Question and Answer

## QuestionPhysicsClass 12

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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