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To solve the circuit problem, we need to determine the voltages across points AC, BC, and the current flowing through \( R_2 \). Let's break down the process step by step:

### 1. Simplifying the Circuit:

- **Resistors \( R_4 \) and \( R_5 \)** are in parallel.
  \[
  \frac{1}{R_{\text{parallel}}} = \frac{1}{R_4} + \frac{1}{R_5} = \frac{1}{100 \, \Omega} + \frac{1}{300 \, \Omega}
  \]
  Solving for \( R_{\text{parallel}} \).

- After finding \( R_{\text{parallel}} \), this resistance is in series with \( R_3 \), forming a new total resistance between points \( A \) and \( C \).

  Total resistance between \( A \) and \( C \) will be:
  \[
  R_{\text{AC}} = R_{\text{parallel}} + R_3
  \]

- **Resistors \( R_1 \) and \( R_2 \)** are in series between points \( B \) and \( C \), so the total resistance between \( B \) and \( C \) is:
  \[
  R_{\text{BC}} = R_1 + R_2
  \]

### 2. Total Circuit Resistance:
The circuit is essentially a combination of two branches (AC and BC) in parallel. The total resistance of the circuit is calculated as:
\[
\frac{1}{R_{\text{total}}} = \frac{1}{R_{\text{AC}}} + \frac{1}{R_{\text{BC}}}
\]

### 3. Finding the Currents:
Using Ohm's law, the total current in the circuit can be found by:
\[
I_{\text{total}} = \frac{V_{\text{total}}}{R_{\text{total}}}
\]
Where \( V_{\text{total}} = 20 \, \text{V} \).

### 4. Current in Each Branch:
Once the total current is found, we can find the current through the \( AC \) and \( BC \) branches using the current divider rule.

### 5. Voltage Across AC and BC:
The voltage across \( AC \) and \( BC \) is calculated using Ohm’s law for each branch:
\[
V_{AC} = I_{\text{AC}} \times R_{\text{AC}}
\]
\[
V_{BC} = I_{\text{BC}} \times R_{\text{BC}}
\]

### 6. Current Through \( R_2 \):
The current through \( R_2 \) is the same as the current in the \( BC \) branch, since \( R_1 \) and \( R_2 \) are in series.

Now let me calculate the values.Here are the calculated values for the given circuit:

1. The equivalent resistance of \( R_4 \) and \( R_5 \) in parallel is \( 75 \, \Omega \).
2. The total resistance in the AC branch is \( 125 \, \Omega \).
3. The total resistance in the BC branch is \( 50 \, \Omega \).
4. The total circuit resistance is \( 35.71 \, \Omega \).
5. The total current in the circuit is approximately \( 0.56 \, \text{A} \).
6. The current in the AC branch is \( 0.16 \, \text{A} \).
7. The current in the BC branch (and through \( R_2 \)) is \( 0.4 \, \text{A} \).
8. The voltage across AC is \( 20 \, \text{V} \).
9. The voltage across BC is also \( 20 \, \text{V} \).

This means that the voltage across both branches is equal, which is consistent since they are connected in parallel.

Do you need more detailed explanations, or do you have any other questions?

### Related Questions:
1. How do you calculate the power dissipated by \( R_2 \)?
2. What would happen to the total current if \( R_4 \) was increased?
3. How would the circuit change if \( R_3 \) was removed?
4. What is the total power consumption of the circuit?
5. How can you verify the current divider rule for this circuit?

### Tip:
Always double-check if resistors are in series or parallel before simplifying the circuit to avoid mistakes in calculations.

Consider the circuit below, determine the voltages across AC, BC and the current flow through R2.

Solve a circuit problem by finding the voltages across points AC and BC, and calculating the current through R2. This problem involves applying Ohm's law and the current divider rule to simplify series and parallel circuits. Suitable for Grades 10-12.

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