In the circuit shown, voltage $$V_o$$ (V) is most nearly:

A. 19.5

B. 18.5

C. 16.5

D. 3.0

We will be using Kirchhoff's voltage and current laws to solve this question.

First, we want to assign polarities to our circuit elements and write out our chosen voltage loops:

In addition, we also want to label our currents into the central node:

Based on these conventions, we can begin deriving equations.

Kirchhoff's Current Law (KCL) states that the sum of all currents entering or leaving a node must be equal to 0. Therefore, we can derive the following equation from our circuit:

${I}_{1}-{I}_{2}-{I}_{3}=0\phantom{\rule{0ex}{0ex}}{I}_{1}={I}_{2}+{I}_{3}(Equation\#1)$

We can also derive another equation from this using Ohm's law:

${I}_{2}=\frac{{V}_{0}}{6\Omega}\phantom{\rule{0ex}{0ex}}{I}_{1}=\frac{{V}_{0}}{6\Omega}+{I}_{3}(Equation\#2)$

Kirchhoff's Voltage Law (KVL) states that the sum of all voltages around any closed loop within the circuit must be equal to 0. Using our two loops that we defined as well as the assigned polarities, we may write the following equations:

$FromLoop1:{V}_{0}-\left(6\Omega \right)\left({I}_{3}\right)-6V=0(Equation\#3)\phantom{\rule{0ex}{0ex}}FromLoop2:36V-\left(3\Omega \right)\left({I}_{1}\right)-\left(6\Omega \right)\left({I}_{3}\right)-6V=0(Equation\#4)$

By simplifying Equation #4, we can find the relationship between ${I}_{1}$ and ${I}_{3}$:

$36V-\left(3\Omega \right)\left({I}_{1}\right)-\left(6\Omega \right)\left({I}_{3}\right)-6V=0\phantom{\rule{0ex}{0ex}}30V-\left(3\Omega \right)\left({I}_{1}\right)-\left(6\Omega \right)\left({I}_{3}\right)=0\phantom{\rule{0ex}{0ex}}30V-\left(3\Omega \right)\left({I}_{1}\right)=\left(6\Omega \right)\left({I}_{3}\right)\phantom{\rule{0ex}{0ex}}5-\frac{{I}_{1}}{2}={I}_{3}(Equation\#5)$

We can then plug this expression in for ${I}_{3}$ in Equation #3:

${V}_{0}-\left(6\Omega \right)\left({I}_{3}\right)-6V=0\phantom{\rule{0ex}{0ex}}{V}_{0}-\left(6\Omega \right)(5-\frac{{I}_{1}}{2})-6V=0\phantom{\rule{0ex}{0ex}}{V}_{0}-(30-3{I}_{1})-6=0\phantom{\rule{0ex}{0ex}}{V}_{0}-36+3{I}_{1}=0\phantom{\rule{0ex}{0ex}}{V}_{0}-36=-3{I}_{1}\phantom{\rule{0ex}{0ex}}\frac{-{V}_{0}}{3}+12={I}_{1}(Equation\#6)$

Finally, we plug in Equations #5 and #6 back into Equation #2 to find our answer:

$(\frac{-{V}_{0}}{3}+12)=\left(\frac{{V}_{0}}{6}\right)+(5-\frac{{I}_{1}}{2})\phantom{\rule{0ex}{0ex}}(\frac{-{V}_{0}}{3}+12)=\left(\frac{{V}_{0}}{6}\right)+(5-\frac{(\frac{-{V}_{0}}{3}+12)}{2})\phantom{\rule{0ex}{0ex}}(\frac{-{V}_{0}}{3}+12)=\left(\frac{{V}_{0}}{6}\right)+(5+\frac{{V}_{0}}{6}-6)\phantom{\rule{0ex}{0ex}}-2{V}_{0}+72={V}_{0}+30+{V}_{0}-36\phantom{\rule{0ex}{0ex}}-4{V}_{0}=-78\phantom{\rule{0ex}{0ex}}{V}_{0}=19.5$

Our answer is therefore A, 19.5.

Copyright © 2024 Savvy Engineer