Question

a) use mesh current to solve for current through the 800ω resistor and v₀. 50 v 80 ω 800 ω v₀ 50 ω 40 ω 750 ma 200 ω

Explanation:

Step1: Define mesh currents

Let the mesh - currents be $I_1$, $I_2$, and $I_3$ in the three meshes (say, left - most, middle, and right - most meshes respectively). The current source of $750\ mA = 0.75\ A$ is in the left - most mesh. Assume $I_1$ flows clock - wise, $I_2$ flows clock - wise, and $I_3$ flows clock - wise.

Step2: Write mesh equations

For the left - most mesh:
The voltage source is $50\ V$. The resistors in this mesh are $50\ \Omega$ and $200\ \Omega$. The current source is $0.75\ A$. Since the current source is in the mesh, we can write the equation considering the voltage drops. The voltage drop across the $50\ \Omega$ resistor due to the current difference between the first and second meshes is $50(I_1 - I_2)$. The voltage drop across the $200\ \Omega$ resistor is $200I_1$. So, $50=50(I_1 - I_2)+200I_1$. Simplifying gives $50 = 250I_1-50I_2$.
For the middle mesh:
The resistors are $50\ \Omega$, $40\ \Omega$, and $80\ \Omega$. The voltage drops are $50(I_2 - I_1)$, $40I_2$, and $80(I_2 - I_3)$. So, $50(I_2 - I_1)+40I_2+80(I_2 - I_3)=0$. Simplifying gives $- 50I_1 + 170I_2-80I_3 = 0$.
For the right - most mesh:
The resistor is $800\ \Omega$ and $80\ \Omega$. The voltage drop across the $80\ \Omega$ resistor is $80(I_3 - I_2)$ and across the $800\ \Omega$ resistor is $800I_3$. So, $80(I_3 - I_2)+800I_3 = 0$. Simplifying gives $-80I_2 + 880I_3 = 0$, or $I_2 = 11I_3$.

Step3: Substitute and solve

Substitute $I_2 = 11I_3$ into $-50I_1 + 170I_2-80I_3 = 0$. We get $-50I_1+170\times11I_3-80I_3 = 0$, or $-50I_1+1870I_3 - 80I_3=0$, or $-50I_1 + 1790I_3 = 0$, or $I_1=\frac{179}{5}I_3$.
Substitute $I_1=\frac{179}{5}I_3$ and $I_2 = 11I_3$ into $50 = 250I_1-50I_2$.
$50=250\times\frac{179}{5}I_3-50\times11I_3$.
$50 = 8950I_3-550I_3$.
$50 = 8400I_3$.
$I_3=\frac{50}{8400}=\frac{1}{168}\ A$.

Step4: Find the current through the $800\ \Omega$ resistor and $v_o$

The current through the $800\ \Omega$ resistor is $I_3=\frac{1}{168}\ A\approx5.95\ mA$.
The voltage $v_o$ across the $800\ \Omega$ resistor is given by Ohm's law $v_o = 800I_3$. Substituting $I_3=\frac{1}{168}\ A$, we get $v_o=\frac{800}{168}=\frac{100}{21}\approx4.76\ V$.

Answer:

The current through the $800\ \Omega$ resistor is $\frac{1}{168}\ A\approx5.95\ mA$, and $v_o=\frac{100}{21}\ V\approx4.76\ V$.