Step-by-step solution for a combined series-parallel circuit problem, showing resistor combinations and calculations.

Problem Analysis and Solution

The given problem involves analyzing a combined series-parallel circuit. We are tasked with finding:

1. The current leaving the battery.
2. The power supplied by the battery.
3. The voltage across \( R_5 \).

#### Step 1: Combine Resistors in Series
The resistors \( R_3 \), \( R_4 \), and \( R_5 \) are in series. The equivalent resistance \( R_{345} \) is calculated as:
\[
R_{345} = R_3 + R_4 + R_5 = 5\, \Omega + 10\, \Omega + 15\, \Omega = 30\, \Omega
\]

#### Step 2: Redraw the Circuit
After combining \( R_3 \), \( R_4 \), and \( R_5 \) into \( R_{345} \), the circuit can be redrawn as follows:
- \( R_1 \) and \( R_2 \) are in parallel.
- \( R_6 \) and \( R_7 \) are in parallel.
- \( R_{345} \) is in series with the parallel combinations.

#### Step 3: Combine Resistors in Parallel
##### (a) Combine \( R_1 \) and \( R_2 \):
\[
R_{12} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2}} = \frac{1}{\frac{1}{20\, \Omega} + \frac{1}{20\, \Omega}} = \frac{1}{\frac{1}{10\, \Omega}} = 10\, \Omega
\]

##### (b) Combine \( R_6 \) and \( R_7 \):
\[
R_{67} = \frac{1}{\frac{1}{R_6} + \frac{1}{R_7}} = \frac{1}{\frac{1}{25\, \Omega} + \frac{1}{100\, \Omega}} = \frac{1}{\frac{4}{100\, \Omega} + \frac{1}{100\, \Omega}} = \frac{1}{\frac{5}{100\, \Omega}} = 20\, \Omega
\]

#### Step 4: Calculate the Total Resistance
Now, the circuit consists of:
- \( R_{12} = 10\, \Omega \)
- \( R_{67} = 20\, \Omega \)
- \( R_{345} = 30\, \Omega \)

These are all in series. The total resistance \( R_{\text{total}} \) is:
\[
R_{\text{total}} = R_{12} + R_{67} + R_{345} = 10\, \Omega + 20\, \Omega + 30\, \Omega = 60\, \Omega
\]

#### Step 5: Calculate the Current Leaving the Battery
Using Ohm's Law:
\[
I_{\text{total}} = \frac{V_{\text{battery}}}{R_{\text{total}}} = \frac{120\, \text{V}}{60\, \Omega} = 2\, \text{A}
\]

#### Step 6: Calculate the Power Supplied by the Battery
The power supplied by the battery is given by:
\[
P_{\text{battery}} = V_{\text{battery}} \times I_{\text{total}} = 120\, \text{V} \times 2\, \text{A} = 240\, \text{W}
\]

#### Step 7: Calculate the Voltage Across \( R_5 \)
Since \( R_3 \), \( R_4 \), and \( R_5 \) are in series, the voltage across \( R_5 \) is part of the total voltage drop across \( R_{345} \). The voltage drop across \( R_{345} \) is:
\[
V_{345} = I_{\text{total}} \times R_{345} = 2\, \text{A} \times 30\, \Omega = 60\, \text{V}
\]

The voltage across \( R_5 \) is proportional to its resistance in the series combination:
\[
V_5 = I_{\text{total}} \times R_5 = 2\, \text{A} \times 15\, \Omega = 30\, \text{V}
\]

Final Answers

\[
\boxed{
\begin{aligned}
&\text{(a) The current leaving the battery: } 2\, \text{A} \\
&\text{(b) The power supplied by the battery: } 240\, \text{W} \\
&\text{(c) The voltage across } R_5: 30\, \text{V}
\end{aligned}
}
\]