An ideal battery sends a current of 5A in a resistor. When another resistor of value $10\Omega$ is connected in parallel, the current through the battery is increased to 6A. Find the resistance of the first resistor.
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$\text{R}_1=\text{R},\text{i}_1=5\text{A}$
$\text{R}_2=\frac{10\text{R}}{10+\text{R}},\text{i}_2=6\text{A}$
Since potential constant,
$\text{i}_1\text{R}_1=\text{i}_2\text{R}_2$
$\Rightarrow5\times\text{R}=\frac{6\times10^\text{R}}{10+\text{R}}$
$\Rightarrow(10+\text{R})5=60$
$\Rightarrow5\text{R}=10\Rightarrow\text{R}=2\Omega.$
$\text{R}_2=\frac{10\text{R}}{10+\text{R}},\text{i}_2=6\text{A}$
Since potential constant,
$\text{i}_1\text{R}_1=\text{i}_2\text{R}_2$
$\Rightarrow5\times\text{R}=\frac{6\times10^\text{R}}{10+\text{R}}$
$\Rightarrow(10+\text{R})5=60$
$\Rightarrow5\text{R}=10\Rightarrow\text{R}=2\Omega.$
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