https://file.aiquickdraw.com/mathbot/file/1726725889897rnatqac3.jpg

To solve this problem, we need to calculate the effective resistance of the appliance using the given information. Here are the steps:

### Given:
- Current (\( I \)) = 2.4 A
- Voltage (\( V \)) = 220 V
- Phase angle (\( \theta \)) = 34 degrees (current lags the voltage by this amount)

The effective resistance (\( R_{\text{eff}} \)) in an AC circuit where the current lags the voltage can be found using the **impedance** and the cosine of the phase angle:

\[
R_{\text{eff}} = \frac{V}{I} \times \cos(\theta)
\]

### Step-by-step solution:

1. **Calculate impedance** (\( Z \)):
   \[
   Z = \frac{V}{I} = \frac{220 \, \text{V}}{2.4 \, \text{A}} = 91.67 \, \Omega
   \]

2. **Calculate effective resistance**:
   Use the formula:
   \[
   R_{\text{eff}} = Z \times \cos(\theta)
   \]
   where \( \theta = 34^\circ \).

   First, calculate \( \cos(34^\circ) \):
   \[
   \cos(34^\circ) \approx 0.829
   \]

   Now, compute the effective resistance:
   \[
   R_{\text{eff}} = 91.67 \, \Omega \times 0.829 = 75.99 \, \Omega
   \]

Thus, the effective resistance is **75.99 ohms**.

Would you like further details or have any questions?

### Related Questions:
1. How do you calculate the impedance in an AC circuit?
2. What is the significance of the phase angle in an AC circuit?
3. How does the power factor relate to the effective resistance?
4. How would the effective resistance change if the phase angle was different?
5. Can you explain the difference between impedance and resistance?

### Tip:
In AC circuits, the **power factor** (\( \cos(\theta) \)) determines how efficiently electrical power is converted into useful work, and it affects the overall effective resistance.

An appliance draws 2.4 ampere of current when plugged to a 220-volt outlet. It was further measured that the current lags the voltage by 34 degrees. What is the effective resistance of the device in ohms? Provide answer in two decimal places.

Learn how to calculate the effective resistance of an AC device where the current lags the voltage by 34 degrees. The appliance draws 2.4 amperes at 220 volts. Suitable for Grades 10-12 physics students.

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