1) write the mesh equations for the following circuit and find $i_r$ based on mesh currents.\n2) using…

1) write the mesh equations for the following circuit and find $i_r$ based on mesh currents.\n2) using superposition, find the voltage $v_1$

1) write the mesh equations for the following circuit and find $i_r$ based on mesh currents.\n2) using superposition, find the voltage $v_1$

Answer

Explanation:

Step1: Define mesh currents for first - circuit

Let the mesh currents be $I_1$ in the left - hand mesh and $I_2$ in the right - hand mesh. The current source of 5 A is common to both meshes. Assume $I_1$ flows clock - wise and $I_2$ flows clock - wise. For the left - hand mesh: The voltage sources are 12 V and 4 V, and the resistors are 2 $\Omega$ and 6 $\Omega$. Using Kirchhoff's Voltage Law (KVL), we have $12-4=(2 + 6)I_1-2I_2$, which simplifies to $8 = 8I_1-2I_2$. For the right - hand mesh: The resistors are 2 $\Omega$, 8 $\Omega$ and 4 $\Omega$. Since there is a current source of 5 A between the meshes, we know that $I_1 - I_2=5$. We can rewrite the first equation as $4 = 4I_1-I_2$ and substitute $I_2=I_1 - 5$ into it. $4 = 4I_1-(I_1 - 5)$ $4 = 4I_1-I_1 + 5$ $3I_1=-1$ $I_1=-\frac{1}{3}$ A and $I_2=-\frac{1}{3}-5=-\frac{16}{3}$ A. The current $I_R$ through the 8 - $\Omega$ resistor is $I_R = I_2=-\frac{16}{3}$ A.

Step2: Use superposition for second - circuit

Case 1: Consider the 12 V voltage source only (replace the 6 A current source with an open - circuit) The total resistance in the circuit is $R_{total}=10 + 9+9=28$ $\Omega$. The current in the circuit $I=\frac{12}{28}=\frac{3}{7}$ A. The voltage across the 9 - $\Omega$ resistor (where $V_1$ is measured) due to the 12 V source, $V_{11}=\frac{3}{7}\times9=\frac{27}{7}$ V. Case 2: Consider the 6 A current source only (replace the 12 V voltage source with a short - circuit) The two 9 - $\Omega$ resistors are in parallel with the 10 - $\Omega$ resistor. The equivalent resistance of the two 9 - $\Omega$ resistors in parallel is $R_{eq9}=\frac{9\times9}{9 + 9}=\frac{9}{2}$ $\Omega$. The total resistance seen by the 6 A current source is $R_{total}=10+\frac{9}{2}=\frac{20 + 9}{2}=\frac{29}{2}$ $\Omega$. The voltage across the parallel combination of 9 - $\Omega$ resistors due to the 6 A source is $V_{12}=6\times\frac{9}{2}=27$ V. By superposition, $V_1 = V_{11}+V_{12}=\frac{27}{7}+27=\frac{27+189}{7}=\frac{216}{7}\approx30.86$ V

Answer:

  1. $I_R=-\frac{16}{3}$ A
  2. $V_1=\frac{216}{7}$ V