A resistor ` R_(1)` consumes electrical power `P_(1)` when connected to an `emf epsilon`. When resistor `R_(2)` is connected to the same `emf` , it consumes electrical power `P_(2)` . In terms of `P_(1) and P_(2)`, what is the total electrical power consumed when they are both connected to this emf source
(a) in parallel
(b) in series

Explanation

To solve the problem, we need to determine the total electrical power consumed when two resistors \( R_1 \) and \( R_2 \) are connected to an EMF \( \epsilon \) in both parallel and series configurations.

### Step 1: Understand the Power Consumption Formula
The power consumed by a resistor when connected to a voltage source is given by the formula:
\[ P = \frac{V^2}{R} \]
where \( P \) is the power, \( V \) is the voltage across the resistor, and \( R \) is the resistance.

### Step 2: Calculate Power in Series
When resistors \( R_1 \) and \( R_2 \) are connected in series, the equivalent resistance \( R_s \) is:
\[ R_s = R_1 + R_2 \]
The power consumed in series \( P_s \) can be expressed as:
\[ P_s = \frac{V^2}{R_s} = \frac{\epsilon^2}{R_1 + R_2} \]

### Step 3: Relate Power to Individual Powers
From the individual powers \( P_1 \) and \( P_2 \) for resistors \( R_1 \) and \( R_2 \):
\[ P_1 = \frac{\epsilon^2}{R_1} \quad \text{and} \quad P_2 = \frac{\epsilon^2}{R_2} \]
Thus, we can express \( \frac{1}{R_1} \) and \( \frac{1}{R_2} \) in terms of \( P_1 \) and \( P_2 \):
\[ \frac{1}{R_1} = \frac{P_1}{\epsilon^2} \quad \text{and} \quad \frac{1}{R_2} = \frac{P_2}{\epsilon^2} \]
Substituting these into the series power formula:
\[ P_s = \frac{\epsilon^2}{R_1 + R_2} = \frac{\epsilon^2}{\left(\frac{\epsilon^2}{P_1} + \frac{\epsilon^2}{P_2}\right)} = \frac{P_1 P_2}{P_1 + P_2} \]

### Step 4: Calculate Power in Parallel
When resistors \( R_1 \) and \( R_2 \) are connected in parallel, the equivalent resistance \( R_p \) is given by:
\[ \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} \]
Thus, the total power consumed in parallel \( P_p \) can be expressed as:
\[ P_p = \frac{\epsilon^2}{R_p} = \epsilon^2 \left( \frac{1}{R_1} + \frac{1}{R_2} \right) \]
Substituting the values of \( \frac{1}{R_1} \) and \( \frac{1}{R_2} \):
\[ P_p = \epsilon^2 \left( \frac{P_1}{\epsilon^2} + \frac{P_2}{\epsilon^2} \right) = P_1 + P_2 \]

### Final Answers
(a) The total electrical power consumed when connected in series is:
\[ P_s = \frac{P_1 P_2}{P_1 + P_2} \]

(b) The total electrical power consumed when connected in parallel is:
\[ P_p = P_1 + P_2 \]