Example of a combined series-parallel circuit with given resistor values and step-by-step solution to find total current, power, and voltage across a specific resistor.

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 of resistors in series is the sum of their individual resistances:
\[
R_{345} = R_3 + R_4 + R_5
\]
Substitute the given values:
\[
R_{345} = 5\, \Omega + 10\, \Omega + 15\, \Omega = 30\, \Omega
\]

#### 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.
- The resulting parallel combinations are then in series with \( R_{345} \).

#### Step 3: Combine Resistors in Parallel
##### (a) Combine \( R_1 \) and \( R_2 \)
The equivalent resistance of two resistors in parallel is given by:
\[
R_{12} = \frac{R_1 \cdot R_2}{R_1 + R_2}
\]
Substitute the given values:
\[
R_{12} = \frac{20\, \Omega \cdot 20\, \Omega}{20\, \Omega + 20\, \Omega} = \frac{400\, \Omega^2}{40\, \Omega} = 10\, \Omega
\]

##### (b) Combine \( R_6 \) and \( R_7 \)
Similarly, for \( R_6 \) and \( R_7 \):
\[
R_{67} = \frac{R_6 \cdot R_7}{R_6 + R_7}
\]
Substitute the given values:
\[
R_{67} = \frac{25\, \Omega \cdot 100\, \Omega}{25\, \Omega + 100\, \Omega} = \frac{2500\, \Omega^2}{125\, \Omega} = 20\, \Omega
\]

#### Step 4: Combine All Resistances
Now, the circuit consists of three resistors in series:
- \( R_{12} = 10\, \Omega \)
- \( R_{67} = 20\, \Omega \)
- \( R_{345} = 30\, \Omega \)

The total resistance \( R_{\text{total}} \) is:
\[
R_{\text{total}} = R_{12} + R_{67} + R_{345}
\]
Substitute the values:
\[
R_{\text{total}} = 10\, \Omega + 20\, \Omega + 30\, \Omega = 60\, \Omega
\]

#### Step 5: Calculate the Current Leaving the Battery
The current \( I_{\text{total}} \) leaving the battery can be found using Ohm's Law:
\[
I_{\text{total}} = \frac{V_{\text{batt}}}{R_{\text{total}}}
\]
Substitute the given values:
\[
I_{\text{total}} = \frac{120\, \text{V}}{60\, \Omega} = 2\, \text{A}
\]

#### Step 6: Calculate the Power Supplied by the Battery
The power \( P \) supplied by the battery is given by:
\[
P = V_{\text{batt}} \cdot I_{\text{total}}
\]
Substitute the values:
\[
P = 120\, \text{V} \cdot 2\, \text{A} = 240\, \text{W}
\]

#### Step 7: Calculate the Voltage Across \( R_5 \)
The voltage across \( R_5 \) is the same as the voltage across \( R_{345} \) because \( R_3 \), \( R_4 \), and \( R_5 \) are in series. The voltage across \( R_{345} \) can be found using Ohm's Law:
\[
V_{345} = I_{\text{total}} \cdot R_{345}
\]
Substitute the values:
\[
V_{345} = 2\, \text{A} \cdot 30\, \Omega = 60\, \text{V}
\]

Since \( R_3 \), \( R_4 \), and \( R_5 \) are in series, the voltage across \( R_5 \) is:
\[
V_5 = I_{\text{total}} \cdot R_5
\]
Substitute the values:
\[
V_5 = 2\, \text{A} \cdot 15\, \Omega = 30\, \text{V}
\]

Final Answers

\[
\boxed{(a)\ 2\, \text{A},\ (b)\ 240\, \text{W},\ (c)\ 30\, \text{V}}
\]