Exercise 27.7 Hints: Getting Started | I'm Stuck What if we have another heater rated at 120 V with power 2.0 kW? (A) What is the resistance of the heater, and the current when it is under normal operation? R = Ω I = A (B) If we connect it a 240 V source, assuming that it still works, what will be the actual power consumption now (in kW)? kW

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**Example 27.7 Power in an Electric Heater**

**Problem:**
An electric heater is constructed by applying a potential difference of 120 V to a Nichrome wire that has a total resistance of 7.00 Ω. Find the current carried by the wire and the power rating of the heater.

**Solution:**

Using the formula \( \Delta V = IR \), we calculate the current:

\[
I = \frac{\Delta V}{R} = \frac{120 \, \text{V}}{7.00 \, \Omega} = \underline{\hspace{1cm}} \, \text{A}
\]

To find the power rating, use the expression \( \mathscr{P} = I^2R \):

\[
\mathscr{P} = I^2(7.00 \, \Omega) = \underline{\hspace{1cm}} \, \text{W}
\]

**What If?**

What if the heater were accidentally connected to a 240 V supply? (This is difficult to do because the shape and orientation of the metal contacts in 240 V plugs are different from those in 120 V plugs.) How would this affect the current carried by the heater and the power rating of the heater?

**Answer:**

If we doubled the applied potential difference, Equation 27.8 tells us that the current would double. According to Equation 27.23, \( \mathscr{P} = (\Delta V)^2/R \), the power would be four times larger.
Transcribed Image Text:**Example 27.7 Power in an Electric Heater** **Problem:** An electric heater is constructed by applying a potential difference of 120 V to a Nichrome wire that has a total resistance of 7.00 Ω. Find the current carried by the wire and the power rating of the heater. **Solution:** Using the formula \( \Delta V = IR \), we calculate the current: \[ I = \frac{\Delta V}{R} = \frac{120 \, \text{V}}{7.00 \, \Omega} = \underline{\hspace{1cm}} \, \text{A} \] To find the power rating, use the expression \( \mathscr{P} = I^2R \): \[ \mathscr{P} = I^2(7.00 \, \Omega) = \underline{\hspace{1cm}} \, \text{W} \] **What If?** What if the heater were accidentally connected to a 240 V supply? (This is difficult to do because the shape and orientation of the metal contacts in 240 V plugs are different from those in 120 V plugs.) How would this affect the current carried by the heater and the power rating of the heater? **Answer:** If we doubled the applied potential difference, Equation 27.8 tells us that the current would double. According to Equation 27.23, \( \mathscr{P} = (\Delta V)^2/R \), the power would be four times larger.
**Exercise 27.7**

**Hints:** Getting Started | I'm Stuck

What if we have another heater rated at 120 V with power 2.0 kW?

(A) What is the resistance of the heater, and the current when it is under normal operation?
- \( R = \) [_____] \(\Omega\)
- \( I = \) [_____] A

(B) If we connect it to a 240 V source, assuming that it still works, what will be the actual power consumption now (in kW)?
- [_____] kW
Transcribed Image Text:**Exercise 27.7** **Hints:** Getting Started | I'm Stuck What if we have another heater rated at 120 V with power 2.0 kW? (A) What is the resistance of the heater, and the current when it is under normal operation? - \( R = \) [_____] \(\Omega\) - \( I = \) [_____] A (B) If we connect it to a 240 V source, assuming that it still works, what will be the actual power consumption now (in kW)? - [_____] kW
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