Question

# We have a $60-\Omega$ resistance, a $20-\Omega$ resistance, and an unknown resistance $R_x$ in parallel with a 15 mA current source. The current through the unknown resistance is 10 mA. Determine the value of $R_x.$

Solution

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Since we know that the current through resistance $R_x$ is 10 mA, sum of currents through brances with 60-$\Omega$ and 20-$\Omega$ resistances is according to Kirchhoff's current law:

$i_{60\Omega}+i_{20\Omega}=15-10=5\ \text{mA}$

Equivalent resistance of parallel comibination of 60-$\Omega$ and 20-$\Omega$ resistances is:

$R_{eq}=\frac{1}{1/60+1/20}=15\ \Omega$

The voltage across parallel is:

$V=(i_{60\Omega}+i_{20\Omega})R_{eq}=5\cdot 15=75\ \text{mA}$

Since we know voltage across resistance $R_x$ and current flowing through it, we can determine value of resistance itself, using Ohm's law:

$\boxed{R_x=\frac{V}{I}=\frac{75}{10}=7.5\ \Omega}$

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