Question

We have a 60Ω60-\Omega resistance, a 20Ω20-\Omega resistance, and an unknown resistance RxR_x in parallel with a 15 mA current source. The current through the unknown resistance is 10 mA. Determine the value of Rx.R_x.

Solution

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

i60Ω+i20Ω=1510=5 mAi_{60\Omega}+i_{20\Omega}=15-10=5\ \text{mA}

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

Req=11/60+1/20=15 ΩR_{eq}=\frac{1}{1/60+1/20}=15\ \Omega

The voltage across parallel is:

V=(i60Ω+i20Ω)Req=515=75 mAV=(i_{60\Omega}+i_{20\Omega})R_{eq}=5\cdot 15=75\ \text{mA}

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

Rx=VI=7510=7.5 Ω\boxed{R_x=\frac{V}{I}=\frac{75}{10}=7.5\ \Omega}

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