To solve the problem, we need to determine the voltage \( V \) across the battery and the resistance \( R_E \). Let's break it down step by step.
Step 1: Determine the current \( I_4 \) through resistor \( R_F \)
From Kirchhoff's Current Law (KCL) at the node where \( I_1 \), \( I_2 \), and \( I_3 \) meet:
\[
I_1 = I_2 + I_3 + I_4
\]
Substitute the given values:
\[
2.5 \, \text{A} = 1.2 \, \text{A} + 1.0 \, \text{A} + I_4
\]
Solve for \( I_4 \):
\[
I_4 = 2.5 \, \text{A} - 1.2 \, \text{A} - 1.0 \, \text{A} = 0.3 \, \text{A}
\]
Step 2: Calculate the voltage drop across \( R_A \)
The voltage drop across \( R_A \) is:
\[
V_{R_A} = I_1 \cdot R_A = 2.5 \, \text{A} \cdot 8 \, \Omega = 20 \, \text{V}
\]
Step 3: Calculate the voltage drop across \( R_B \) and \( R_C \)
The voltage drop across \( R_B \) is:
\[
V_{R_B} = I_2 \cdot R_B = 1.2 \, \text{A} \cdot 5 \, \Omega = 6 \, \text{V}
\]
The voltage drop across \( R_C \) is:
\[
V_{R_C} = I_3 \cdot R_C = 1.0 \, \text{A} \cdot 6 \, \Omega = 6 \, \text{V}
\]
Step 4: Determine the voltage drop across \( R_D \) and \( R_E \)
Since \( R_B \) and \( R_C \) are in parallel with \( R_D \) and \( R_E \), the voltage across \( R_D \) and \( R_E \) must be the same as the voltage across \( R_B \) and \( R_C \):
\[
V_{R_D} + V_{R_E} = V_{R_B} = 6 \, \text{V}
\]
The current through \( R_D \) is:
\[
I_D = \frac{V_{R_D}}{R_D} = \frac{6 \, \text{V}}{7 \, \Omega} = \frac{6}{7} \, \text{A}
\]
The current through \( R_E \) is:
\[
I_E = I_3 - I_D = 1.0 \, \text{A} - \frac{6}{7} \, \text{A} = \frac{7}{7} \, \text{A} - \frac{6}{7} \, \text{A} = \frac{1}{7} \, \text{A}
\]
The voltage drop across \( R_E \) is:
\[
V_{R_E} = I_E \cdot R_E
\]
Since \( V_{R_D} + V_{R_E} = 6 \, \text{V} \):
\[
V_{R_E} = 6 \, \text{V} - V_{R_D} = 6 \, \text{V} - \left( \frac{6}{7} \, \text{A} \cdot 7 \, \Omega \right) = 6 \, \text{V} - 6 \, \text{V} = 0 \, \text{V}
\]
This indicates a mistake in the calculation. Let's re-evaluate \( R_E \):
Step 5: Correct calculation of \( R_E \)
Using the correct approach:
\[
V_{R_E} = I_E \cdot R_E
\]
\[
6 \, \text{V} = \left( \frac{1}{7} \, \text{A} \right) \cdot R_E
\]
Solve for \( R_E \):
\[
R_E = \frac{6 \, \text{V}}{\frac{1}{7} \, \text{A}} = 6 \, \text{V} \cdot 7 = 42 \, \Omega
\]
Step 6: Calculate the total voltage \( V \)
The total voltage \( V \) is the sum of the voltage drops across \( R_A \), \( R_B \), and \( R_F \):
\[
V = V_{R_A} + V_{R_B} + V_{R_F}
\]
The voltage drop across \( R_F \) is:
\[
V_{R_F} = I_4 \cdot R_F = 0.3 \, \text{A} \cdot 9 \, \Omega = 2.7 \, \text{V}
\]
Thus:
\[
V = 20 \, \text{V} + 6 \, \text{V} + 2.7 \, \text{V} = 28.7 \, \text{V}
\]
Final Answers:
\[
\boxed{28.7 \, \text{V}, 42 \, \Omega}
\]
Parent Tip: Review the logic above to help your child master the concept of ohms law practice worksheet.
