https://file.aiquickdraw.com/mathbot/file/17250031713021shvx0z3.jpg

For each of the circuits, find the current I and compute the power absorbed by the resistor.

Let's analyze each circuit in the image to find the current \( I \) and the power absorbed by the resistor.

### Circuit Analysis:
Each circuit has a voltage source connected in series with a resistor of \( 10 \, \text{k}\Omega \). The current \( I \) can be found using Ohm's Law:

\[
I = \frac{V}{R}
\]

where \( V \) is the voltage across the resistor and \( R \) is the resistance.

The power absorbed by the resistor can be calculated using the formula:

\[
P = I^2 R
\]

Let's calculate for each circuit:

1. **First Circuit (Top Left):**
   - Voltage, \( V = 5 \, \text{V} \)
   - Resistance, \( R = 10 \, \text{k}\Omega = 10,000 \, \Omega \)
   - Current, \( I = \frac{5 \, \text{V}}{10,000 \, \Omega} = 0.0005 \, \text{A} = 0.5 \, \text{mA} \)
   - Power, \( P = (0.0005 \, \text{A})^2 \times 10,000 \, \Omega = 0.0025 \, \text{W} = 2.5 \, \text{mW} \)

2. **Second Circuit (Top Right):**
   - Voltage, \( V = 5 \, \text{V} \)
   - Resistance, \( R = 10 \, \text{k}\Omega = 10,000 \, \Omega \)
   - Current, \( I = \frac{5 \, \text{V}}{10,000 \, \Omega} = 0.0005 \, \text{A} = 0.5 \, \text{mA} \)
   - Power, \( P = (0.0005 \, \text{A})^2 \times 10,000 \, \Omega = 0.0025 \, \text{W} = 2.5 \, \text{mW} \)

3. **Third Circuit (Bottom Left):**
   - Voltage, \( V = V \, \text{V} \)
   - Resistance, \( R = 10 \, \text{k}\Omega = 10,000 \, \Omega \)
   - Current, \( I = \frac{V \, \text{V}}{10,000 \, \Omega} = \frac{V}{10,000} \, \text{A} \)
   - Power, \( P = \left(\frac{V}{10,000}\right)^2 \times 10,000 \, \Omega = \frac{V^2}{10,000} \, \text{W} \)

   Since the voltage is denoted as \( V \), the power will depend on the specific value of \( V \).

4. **Fourth Circuit (Bottom Right):**
   - Voltage, \( V = -5 \, \text{V} \)
   - Resistance, \( R = 10 \, \text{k}\Omega = 10,000 \, \Omega \)
   - Current, \( I = \frac{-5 \, \text{V}}{10,000 \, \Omega} = -0.0005 \, \text{A} = -0.5 \, \text{mA} \)
   - Power, \( P = (-0.0005 \, \text{A})^2 \times 10,000 \, \Omega = 0.0025 \, \text{W} = 2.5 \, \text{mW} \)

### Summary:
- **Top Left Circuit:**
  - Current \( I = 0.5 \, \text{mA} \)
  - Power \( P = 2.5 \, \text{mW} \)
  
- **Top Right Circuit:**
  - Current \( I = 0.5 \, \text{mA} \)
  - Power \( P = 2.5 \, \text{mW} \)
  
- **Bottom Left Circuit:**
  - Current \( I = \frac{V}{10,000} \, \text{A} \)
  - Power \( P = \frac{V^2}{10,000} \, \text{W} \)
  
- **Bottom Right Circuit:**
  - Current \( I = -0.5 \, \text{mA} \)
  - Power \( P = 2.5 \, \text{mW} \)

### Do you want details or have any questions?

Here are five related questions:
1. How does the polarity of voltage affect the direction of current in these circuits?
2. How would the current change if the resistance was doubled?
3. What would happen to the power absorbed if the voltage was halved?
4. How does the power dissipated relate to the square of the current?
5. Could you analyze a circuit with multiple resistors in series and parallel?

**Tip:** Always check the polarity of voltage sources before applying Ohm's Law to determine the correct direction of current flow.

Learn how to find the current and calculate the power absorbed by resistors in series circuits using Ohm's Law and the power formula. Detailed analysis of voltage sources and resistances included.

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