Chapter 21, Problem 50P

For the circuit shown in Figure P21.50, we wish to find the currents I₁, I₂, and I₃. Use Kirchhoff’s rules to obtain equations for (a) the upper loop, (b) the lower loop, and (c) the junction on the left side. In each case, suppress units for clarity and simplify, combining the terms. (d) Solve the junction equation for I₃. (e) Using the equation found in part (d), eliminate I₃ from the equation found in part (b). (f) Solve the equations found in parts (a) and (e) simultaneously for the two unknowns I₁ and I₂. (g) Substitute the answers found in part (f) into the junction equation found in part (d), solving for I₃. (h) What is the significance of the negative answer for I₂?

Figure P21.50

To determine

The equations for the upper loop in the circuit diagram.

Answer to Problem 50P

The equation for the upper loop is $\underset{\_}{13.0I_1+18.0I_2=30.0}$.

Explanation of Solution

Write the expression for the Kirchhoff’s loop rule for the upper loop going counter clockwise.

    $\begin{array}{c}-\left(11.0\Omega \right)I_2\text{+12}\text{.0}\text{V}-\left(7.00\Omega \right)I_2-\left(5.00\Omega \right)I_1+18.0\text{V}-\left(8.00\Omega \right)I_1=0\\ 30.0\text{V}-\left(13.00\Omega \right)I_1-\left(18.0\Omega \right)I_2=0\\ \left(13.00\Omega \right)I_1+\left(18.0\Omega \right)I_2=30.0\text{V}\end{array}$        (I)

Conclusion:

Therefore, the equation for the upper loop is $\underset{\_}{13.0I_1+18.0I_2=30.0}$.

To determine

The equations for the lower loop in the circuit diagram.

Answer to Problem 50P

The equation for the lower loop is $\underset{\_}{18.0I_2-5.00I_3=-24.0}$.

Explanation of Solution

Write the expression for the Kirchhoff’s loop rule for the lower loop going counter clockwise.

    $\begin{array}{c}-\left(5.00\Omega \right)I_3+36.0\text{V}+\left(7.00\Omega \right)I_2-12.0\text{V}+\left(11.0\Omega \right)I_2=0\\ 24.0\text{V}+\left(18.0\Omega \right)I_2-\left(5.00\Omega \right)I_3=0\\ \left(18.0\Omega \right)I_2-\left(5.00\Omega \right)I_3=-24.0\text{V}\end{array}$        (II)

Conclusion:

Therefore, the equation for the lower loop is $\underset{\_}{18.0I_2-5.00I_3=-24.0}$.

To determine

The equation of the junction on the left side in the circuit diagram.

Answer to Problem 50P

The equation of the junction in the left side is $\underset{\_}{I_1-I_2-I_3=0}$.

Explanation of Solution

Write the expression for the Kirchhoff’s junction rule for the junction on the left side in the circuit.

    $I_1-I_2-I_3=0$        (III)

Conclusion:

Therefore, the equation of the junction in the left side is $\underset{\_}{I_1-I_2-I_3=0}$.

To determine

The equation for $I_3$ in the circuit.

Answer to Problem 50P

The equation for $I_3$ in the circuit is $\underset{\_}{I_3=I_1-I_2}$.

Explanation of Solution

Use equation (III) to solve for $I_3$.

    $I_3=I_1-I_2$        (IV)

Conclusion:

Therefore, the equation for $I_3$ in the circuit is $\underset{\_}{I_3=I_1-I_2}$.

To determine

The equation of the lower loop without using $I_3$.

Answer to Problem 50P

The equation of the lower loop without using $I_3$ $\underset{\_}{5.00I_1-23.0I_2=24.0}$.

Explanation of Solution

Use equation (IV) in (II) to solve for the lower without using $I_3$.

    $\begin{array}{c}\left(18.0\Omega \right)I_2-\left(5.00\Omega \right)\left(I_1-I_2\right)=-24.0\text{V}\\ \left(\text{23}\text{.0}\Omega \right)I_2-\left(5.00\Omega \right)I_1=-24.0\text{V}\\ \left(5.00\Omega \right)I_1-\left(\text{23}\text{.0}\Omega \right)I_2=24.0\text{V}\end{array}$        (V)

Conclusion:

Therefore, the equation of the lower loop without using $I_3$ $\underset{\_}{5.00I_1-23.0I_2=24.0}$.

To determine

The value of current $I_1$ and $I_2$.

Answer to Problem 50P

The value of current $I_1$ is $\underset{\_}{2.88\text{A}}$ and $I_2$ is $\underset{\_}{-0.416\text{A}}$.

Explanation of Solution

Use equation (V) to solve for $I_1$.

    $I_1=\frac{\left(24.0\text{V}+\left(23.0\Omega \right)I_2\right)}{5.00\Omega }$        (VI)

Use equation (VI) in (I) to solve for $I_2$.

    $\begin{array}{c}\left(13.0\Omega \right)\left[\frac{\left(24.0\text{V}+\left(23.0\Omega \right)I_2\right)}{5.00\Omega }\right]+\left(18.0\Omega \right)I_2=30.0\text{V}\\ \left(13.0\Omega \right)\left(24.0\text{V}+\left(23.0\Omega \right)I_2\right)+\left(5.00\Omega \right)\left(18.0\Omega \right)I_2=\left(5.00\Omega \right)\left(30.0\text{V}\right)\\ \left(389\Omega \right)I_2=-162\text{V}\\ I_2=-0.416\text{A}\end{array}$        (VII)

Use equation (VII) in (I) to solve for $I_1$.

    $\begin{array}{c}\left(13.0\Omega \right)I_1+\left(18.0\Omega \right)\left(-0.416\text{A}\right)=30.0\text{V}\\ \left(13.0\Omega \right)I_1-\left(7.488\text{V}\right)=30.0\text{V}\\ I_1=\frac{\left(30.0\text{V}-7.488\text{V}\right)}{13.0\Omega }\\ =2.88\text{A}\end{array}$        (VIII)

Conclusion:

Therefore, the value of current $I_1$ is $\underset{\_}{2.88\text{A}}$ and $I_2$ is $\underset{\_}{-0.416\text{A}}$.

To determine

The value of current $I_3$.

Answer to Problem 50P

The value of current $I_3$ is $\underset{\_}{3.30\text{A}}$.

Explanation of Solution

Use equation (VIII) and (VII) in (IV) to solve for $I_3$.

    $\begin{array}{c}{I}_3=2.88\text{A}-\left(-0.416\text{A}\right)\\ =3.30\text{A}\end{array}$        (IX)

Conclusion:

Therefore, the value of current $I_3$ is $\underset{\_}{3.30\text{A}}$.

To determine

The significance of negative sign in the current $I_2$.

Answer to Problem 50P

The negative sign for $I_2$ signifies that the direction of the current is opposite the direction shown in figure $1$.

Explanation of Solution

The negative sign in the value of the current $I_2$ signifies that the current flows in the opposite direction as shown in the circuit diagram. The magnitude of the current flowing through the middle branch of the circuit diagram is $0.416\text{A}$ from right to left.

Conclusion:

Therefore, the negative sign for $I_2$ signifies that the direction of the current is opposite the direction shown in figure $1$.