Calculate the current using KVL and KCL for the electrical circuit including the current, voltage, and resistance.

Steps to Solve

1. Redraw and Label the Circuit

First, redraw the circuit diagram to make it clearer and label all the components and nodes. I think the $7 \Omega$ resistor is actually $8 \Omega$. We know that is the case because there is another $7 \Omega$ label which is the current we need to find. The circuit contains:

• A 20V voltage source
• A 60V voltage source
• Three resistors: $6 \Omega$, $8 \Omega$, and $9 \Omega$
• The goal is to find the current $I_2$ flowing through what looks like an $8 \Omega$ resistor but it represents a current that you will find. Since the initial writing is not clear let's rename it to $I_7$
• Two other currents are labeled $I_1$ and $I_2$; however, it appears that $I_2$ is actually $I_7$.

2. Apply KCL at Node A

Apply Kirchhoff's Current Law (KCL) at Node A. KCL states that the sum of currents entering a node must equal the sum of currents leaving the node. $$ I_1 = I_7 + I_2$$

3. Apply KVL in Loop 1

Apply Kirchhoff's Voltage Law (KVL) to the loop containing the 20V source, the $6\Omega$ resistor, and the $8\Omega$ resistor. $$ 20 - 6I_1 - 8I_2 = 0$$

4. Apply KVL in Loop 2

Apply Kirchhoff's Voltage Law (KVL) to the loop containing the $8\Omega$ resistor, the $9\Omega$ resistor, and the 60V source $$ -8I_2 + 9I_7 + 60 = 0$$

5. Express $I_1$ in terms of $I_2$ from Loop 1 KVL

From the KVL equation for Loop 1, express (I_1) in terms of (I_2): $$6I_1 = 20 - 8I_2$$ $$I_1 = \frac{20 - 8I_2}{6} = \frac{10 - 4I_2}{3}$$

6. Express $I_7$ in terms of $I_2$ from Loop 2 KVL

From the KVL equation for Loop 2, express (I_7) in terms of (I_2): $$9I_7 = 8I_2 - 60$$ $$I_7 = \frac{8I_2 - 60}{9}$$

7. Substitute $I_1$ and $I_7$ into the KCL equation

Substitute the expressions for $I_1$ and $I_7$ into the KCL equation from Node A: $$ \frac{10 - 4I_2}{3} = \frac{8I_2 - 60}{9} + I_2 $$

8. Solve for $I_2$

Multiply through by 9 to eliminate fractions: $$ 3(10 - 4I_2) = (8I_2 - 60) + 9I_2 $$ $$ 30 - 12I_2 = 8I_2 - 60 + 9I_2 $$ $$ 30 - 12I_2 = 17I_2 - 60 $$ $$ 90 = 29I_2 $$ $$ I_2 = \frac{90}{29} $$

9. Solve for $I_7$

Using the solution for $I_2$, we can now calculate the final answer $I_7$ $$I_7 = (\frac{8I_2 - 60}{9})$$ $$I_7 = (\frac{8(\frac{90}{29}) - 60}{9})$$ $$I_7 = (\frac{\frac{720}{29} - \frac{1740}{29}}{9})$$ $$I_7 = (\frac{\frac{-1020}{29}}{9})$$ $$I_7 = \frac{-1020}{29*9} = \frac{-1020}{261} = \frac{-340}{87} \approx -3.908 A$$