Question

17. Determine the currents in the network given in the figure.

• A. Current through the 40V battery and the first 40 $\Omega$ resistor is $0.55 A$. Current through the 30 $\Omega$ resistor is $0.6 A$. Current through the 20V battery and the second 40 $\Omega$ resistor is $0.05 A$.
• B. Current through the 40V battery and the first 40 $\Omega$ resistor is $0.05 A$. Current through the 30 $\Omega$ resistor is $0.6 A$. Current through the 20V battery and the second 40 $\Omega$ resistor is $0.55 A$.
• C. Current through the 40V battery and the first 40 $\Omega$ resistor is $0.55 A$. Current through the 30 $\Omega$ resistor is $0.05 A$. Current through the 20V battery and the second 40 $\Omega$ resistor is $0.6 A$.
• D. Current through the 40V battery and the first 40 $\Omega$ resistor is $0.6 A$. Current through the 30 $\Omega$ resistor is $0.55 A$. Current through the 20V battery and the second 40 $\Omega$ resistor is $0.05 A$.

Answer 1

Answer: (A)

The currents in the network are:

• Current through the 40V battery and the first 40 $\Omega$ resistor is $0.55 A$.
• Current through the 30 $\Omega$ resistor is $0.6 A$.
• Current through the 20V battery and the second 40 $\Omega$ resistor is $0.05 A$.

Answer 2

The problem requires us to determine the currents in the given electrical network using Kirchhoff's laws.

1. Circuit Diagram and Current Definitions: Let $I_1$ be the current flowing out of the positive terminal of the 40V battery, through the 40 $\Omega$ resistor. Let $I_3$ be the current flowing out of the positive terminal of the 20V battery, through the 40 $\Omega$ resistor. Let $I_2$ be the current flowing downwards through the 30 $\Omega$ resistor.

Applying Kirchhoff's Current Law (KCL) at the junction where the three branches meet: $I_1 + I_3 = I_2$ (Equation 1)

2. Applying Kirchhoff's Voltage Law (KVL):

• Left Loop: Traversing clockwise from the negative terminal of the 40V battery: $+40V - 40 \Omega \cdot I_1 - 30 \Omega \cdot I_2 = 0$ $40 I_1 + 30 I_2 = 40$ Dividing by 10: $4 I_1 + 3 I_2 = 4 \quad (\text{Equation 2})$

• Right Loop: Traversing clockwise from the negative terminal of the 20V battery: $+20V - 40 \Omega \cdot I_3 - 30 \Omega \cdot I_2 = 0$ $40 I_3 + 30 I_2 = 20$ Dividing by 10: $4 I_3 + 3 I_2 = 2 \quad (\text{Equation 3})$

3. Solving the System of Equations: We have the system:

1. $I_1 + I_3 = I_2$
2. $4 I_1 + 3 I_2 = 4$
3. $4 I_3 + 3 I_2 = 2$

From Equation 1, $I_1 = I_2 - I_3$. Substitute into Equation 2: $4 (I_2 - I_3) + 3 I_2 = 4$ $7 I_2 - 4 I_3 = 4 \quad (\text{Equation 4})$

Now we solve the system formed by Equation 3 and Equation 4: 3. $3 I_2 + 4 I_3 = 2$ 4. $7 I_2 - 4 I_3 = 4$

Adding Equation 3 and Equation 4: $10 I_2 = 6$ $I_2 = 0.6 A$

Substitute $I_2$ into Equation 3: $3 (0.6) + 4 I_3 = 2$ $1.8 + 4 I_3 = 2$ $4 I_3 = 0.2$ $I_3 = 0.05 A$

Substitute $I_2$ and $I_3$ into Equation 1: $I_1 = 0.6 A - 0.05 A$ $I_1 = 0.55 A$

4. Results: The currents are $I_1 = 0.55 A$, $I_2 = 0.6 A$, and $I_3 = 0.05 A$. All currents are positive, indicating the assumed directions are correct.