Question

a) use mesh current to solve for current through the 800ω resistor and vₒ. 800 ω - vₒ + 80 ω 40 ω 50 v + - 50 ω 750 ma 200 ω

Explanation:

Step1: Define mesh currents

Let the mesh - currents be $I_1$ in the left - hand mesh and $I_2$ in the right - hand mesh. Assume $I_1$ is clockwise and $I_2$ is clockwise.

Step2: Write KVL equations

For the left - hand mesh:
$50=80I_1 + 50(I_1 - I_2)+800(I_1 - I_2)$
$50=(80 + 50+800)I_1-(50 + 800)I_2$
$50 = 930I_1-850I_2$

For the right - hand mesh:
The current source of $750\ mA = 0.75\ A$ is in the mesh. We can write the equation considering the voltage drops.
$0 = 800(I_2 - I_1)+40I_2+200I_2$
$0=-800I_1+(800 + 40+200)I_2$
$0=-800I_1 + 1040I_2$
$800I_1=1040I_2$
$I_1=\frac{1040}{800}I_2=\frac{13}{10}I_2$

Step3: Substitute $I_1$ into the first KVL equation

Substitute $I_1=\frac{13}{10}I_2$ into $50 = 930I_1-850I_2$:
$50=930\times\frac{13}{10}I_2-850I_2$
$50=(93\times13)I_2-850I_2$
$50 = 1209I_2-850I_2$
$50 = 359I_2$
$I_2=\frac{50}{359}\ A$
$I_1=\frac{13}{10}\times\frac{50}{359}=\frac{65}{359}\ A$

Step4: Calculate the current through the $800\ \Omega$ resistor

The current through the $800\ \Omega$ resistor, $I = I_1 - I_2$
$I=\frac{65}{359}-\frac{50}{359}=\frac{15}{359}\ A\approx0.0418\ A = 41.8\ mA$

Step5: Calculate $v_o$

Using Ohm's law, $v_o = 800I$
$v_o=800\times\frac{15}{359}\ V\approx33.4\ V$

Answer:

The current through the $800\ \Omega$ resistor is $\frac{15}{359}\ A\approx41.8\ mA$ and $v_o\approx33.4\ V$