https://file.aiquickdraw.com/mathbot/file/17250032532196tyu4k5d.jpg

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

To solve the problem for each of the circuits depicted in the image, we need to follow these steps:

1. **Calculate the current \( I \) using Ohm's Law**:
   \[
   I = \frac{V}{R}
   \]
   where \( V \) is the voltage across the resistor, and \( R \) is the resistance.

2. **Compute the power absorbed by the resistor**:
   \[
   P = I^2 \cdot R
   \]
   Alternatively, you can use:
   \[
   P = \frac{V^2}{R}
   \]

### Analyzing Each Circuit:

#### 1. Top Left Circuit:
- **Voltage Source**: \( V = 5 \, \text{V} \)
- **Resistance**: \( R = 10 \, \text{k}\Omega = 10,000 \, \Omega \)

**Current Calculation**:
\[
I = \frac{5 \, \text{V}}{10,000 \, \Omega} = 0.0005 \, \text{A} = 0.5 \, \text{mA}
\]

**Power Absorbed by the Resistor**:
\[
P = I^2 \cdot R = (0.0005 \, \text{A})^2 \cdot 10,000 \, \Omega = 0.0025 \, \text{W} = 2.5 \, \text{mW}
\]

#### 2. Top Right Circuit:
- **Voltage Source**: \( V = 5 \, \text{V} \)
- **Resistance**: \( R = 10 \, \text{k}\Omega \)

**Current Calculation**:
\[
I = \frac{5 \, \text{V}}{10,000 \, \Omega} = 0.5 \, \text{mA}
\]

**Power Absorbed by the Resistor**:
\[
P = 2.5 \, \text{mW}
\]

(Note: This circuit is identical to the first one.)

#### 3. Bottom Left Circuit:
- **Voltage Source**: \( V = V \) (assuming \( V \) is 5V similar to the other circuits for calculation)
- **Resistance**: \( R = 10 \, \text{k}\Omega \)

**Current Calculation**:
\[
I = \frac{V}{10,000 \, \Omega} 
\]

Assuming \( V = 5 \, \text{V} \):
\[
I = 0.5 \, \text{mA}
\]

**Power Absorbed by the Resistor**:
\[
P = 2.5 \, \text{mW}
\]

(Note: The actual value of \( V \) should be provided for precise calculation.)

#### 4. Bottom Right Circuit:
- **Voltage Source**: \( V = -5 \, \text{V} \)
- **Resistance**: \( R = 10 \, \text{k}\Omega \)

**Current Calculation**:
\[
I = \frac{-5 \, \text{V}}{10,000 \, \Omega} = -0.5 \, \text{mA}
\]

(The negative sign indicates the direction of current opposite to the assumed direction.)

**Power Absorbed by the Resistor**:
\[
P = I^2 \cdot R = 2.5 \, \text{mW}
\]

(Note: Power is always positive as it represents the energy absorbed, irrespective of the current direction.)

### Summary:
- **Top Left Circuit**: \( I = 0.5 \, \text{mA} \), \( P = 2.5 \, \text{mW} \)
- **Top Right Circuit**: \( I = 0.5 \, \text{mA} \), \( P = 2.5 \, \text{mW} \)
- **Bottom Left Circuit**: \( I = \frac{V}{10,000 \, \Omega} \), \( P = \frac{V^2}{10,000 \, \Omega} \) (assuming \( V \) is 5V, results will match the above)
- **Bottom Right Circuit**: \( I = -0.5 \, \text{mA} \), \( P = 2.5 \, \text{mW} \)

Would you like further details or have any specific questions?

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### Follow-up Questions:
1. How does reversing the voltage polarity affect the current direction in a simple resistive circuit?
2. What is the significance of the power sign in AC circuits compared to DC circuits?
3. How would the current change if the resistor's value was halved in any of these circuits?
4. How do different resistor values affect the overall power dissipation in a series circuit?
5. What would happen to the power absorbed if the voltage were doubled?

### Tip:
Always double-check the direction of current and polarity of the voltage source when solving circuits; it helps in correctly interpreting the sign of the current and power.

Learn how to calculate current (I) using Ohm's Law and compute the power absorbed by a resistor in various resistive circuits. Includes detailed steps and examples for different configurations. Suitable for advanced high school and college students.

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