Example of Power Dissipation in Series Circuit with \( R_1 = 150 \, \Omega \) and \( R_2 = 180 \, \Omega \)

See below the calculations for the example of a 150 Ω resistor in series with a 180 Ω resistor, using a total voltage of 10 V. The total resistance, current, voltage drop across each resistor, and power dissipated by each resistor.

Let's calculate the total resistance and the power dissipated by each resistor when \( R_1 = 150 \, \Omega \) and \( R_2 = 180 \, \Omega \), with a total voltage \( V_{total} \).

1. Total Resistance Calculation: For two resistors, \( R_1 = 150 \, \Omega \) and \( R_2 = 180 \, \Omega \), in series, the total resistance \( R_{total} \) is given by:
   \[ R_{total} = R_1 + R_2 \]
   \[ R_{total} = 150 \, \Omega + 180 \, \Omega = 330 \, \Omega \]
2. Current Through the Circuit: The current \( I \) through the series circuit is the same for both resistors and is calculated using Ohm's law:
   \[ I = \frac{V_{total}}{R_{total}} \]
3. Voltage Drop Across Each Resistor: The voltage drop \( V_1 \) across \( R_1 \) and \( V_2 \) across \( R_2 \) can be calculated as:
   \[ V_1 = I \times R_1 \quad \text{and} \quad V_2 = I \times R_2 \]
4. Power Dissipated by Each Resistor: The power dissipated by each resistor, \( P_1 \) for \( R_1 \) and \( P_2 \) for \( R_2 \), can be calculated as:
   \[ P_1 = I^2 \times R_1 \quad \text{and} \quad P_2 = I^2 \times R_2 \]
5. Example Calculation: Assume the total voltage \( V_{total} = 10 \, \text{V} \).
   \[ I = \frac{10 \, \text{V}}{330 \, \Omega} \approx 0.0303 \, \text{A} \]
   \[ V_1 = 0.0303 \, \text{A} \times 150 \, \Omega \approx 4.545 \, \text{V} \]
   \[ V_2 = 0.0303 \, \text{A} \times 180 \, \Omega \approx 5.455 \, \text{V} \]
   \[ P_1 = (0.0303 \, \text{A})^2 \times 150 \, \Omega \approx 0.1377 \, \text{W} \]
   \[ P_2 = (0.0303 \, \text{A})^2 \times 180 \, \Omega \approx 0.1652 \, \text{W} \]