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Let's go through the steps to solve the two questions in this circuit problem:

### 1. Find the current through the battery.
The circuit is composed of multiple resistors, which can be simplified using series and parallel rules.

#### Step 1: Combine \( R_1 \) and \( R_2 \) (top branch)
\( R_1 = 10 \, \Omega \) and \( R_2 = 5 \, \Omega \) are in series, so the total resistance of this branch is:
\[
R_{12} = R_1 + R_2 = 10 \, \Omega + 5 \, \Omega = 15 \, \Omega
\]

#### Step 2: Combine \( R_3 \), \( R_4 \), and \( R_5 \) (bottom branch)
- \( R_3 = 10 \, \Omega \)
- \( R_4 = 5 \, \Omega \)
- \( R_5 = 20 \, \Omega \)

\( R_3 \) and \( R_5 \) are in series, so their total resistance is:
\[
R_{35} = R_3 + R_5 = 10 \, \Omega + 20 \, \Omega = 30 \, \Omega
\]
Now, \( R_{35} \) is in parallel with \( R_4 \), so the equivalent resistance for the bottom branch is:
\[
\frac{1}{R_{eq}} = \frac{1}{R_{35}} + \frac{1}{R_4} = \frac{1}{30} + \frac{1}{5}
\]
\[
\frac{1}{R_{eq}} = \frac{1 + 6}{30} = \frac{7}{30}
\]
\[
R_{eq} = \frac{30}{7} \approx 4.29 \, \Omega
\]

#### Step 3: Find the total resistance
Now, the equivalent resistances of the two branches are in series:
\[
R_{total} = R_{12} + R_{eq} = 15 \, \Omega + 4.29 \, \Omega \approx 19.29 \, \Omega
\]

#### Step 4: Calculate the current through the battery
Using Ohm's law \( V = IR \), where \( V = 10 \, \text{V} \) and \( R_{total} = 19.29 \, \Omega \):
\[
I_{battery} = \frac{V}{R_{total}} = \frac{10 \, \text{V}}{19.29 \, \Omega} \approx 0.518 \, \text{A}
\]

### 2. What is the current through the 20-Ω resistor?
The 20-Ω resistor is \( R_5 \) in the bottom branch. We already know that the equivalent resistance of the bottom branch is \( 4.29 \, \Omega \), and the current splits between the two parallel branches.

#### Step 1: Find the voltage across the parallel section
The voltage across the parallel resistors is the same, and it's equal to the voltage drop across \( R_{12} \), calculated using \( V = IR \):
\[
V_{parallel} = I_{battery} \times R_{12} = 0.518 \, \text{A} \times 15 \, \Omega \approx 7.77 \, \text{V}
\]

#### Step 2: Calculate the current through \( R_{35} \)
The voltage across \( R_{35} \) is also \( 7.77 \, \text{V} \), and the total resistance of \( R_{35} \) is 30 Ω. Using Ohm's law:
\[
I_{R_{35}} = \frac{V_{parallel}}{R_{35}} = \frac{7.77 \, \text{V}}{30 \, \Omega} \approx 0.259 \, \text{A}
\]

This current flows through both \( R_3 \) and \( R_5 \), so the current through the 20-Ω resistor is also \( 0.259 \, \text{A} \).

### Summary of Answers:
1. **Current through the battery:** \( 0.518 \, \text{A} \)
2. **Current through the 20-Ω resistor:** \( 0.259 \, \text{A} \)

Would you like further details or clarifications on any step?

### Related Questions:
1. How would the total current change if \( R_3 \) were increased to 20 Ω?
2. What would happen to the current through \( R_5 \) if the battery voltage increased to 20 V?
3. How can the power dissipated by the 20-Ω resistor be calculated?
4. How would you calculate the total power supplied by the battery?
5. If the resistors were rearranged differently, how would it affect the total current?

### Tip:
For circuits involving series and parallel combinations, always simplify step-by-step, and remember that current is the same through series elements, while voltage is the same across parallel elements.

1. Find the current through the battery. 2. What is the current through the 20-Ω resistor?

Learn how to calculate the current through a battery and a 20-Ω resistor in a complex circuit with series and parallel resistances. This step-by-step guide explains how to apply Ohm's Law and simplify the circuit, ideal for students in Grades 10-12.

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