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
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
College Physics
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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9974cb88-485f-4954-981d-1d67477878d4%2F6040debf-8037-4d26-9499-472c69f1e6b0%2Fxs4e7zz_processed.png&w=3840&q=75)
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9974cb88-485f-4954-981d-1d67477878d4%2F6040debf-8037-4d26-9499-472c69f1e6b0%2Fev8kww1_processed.png&w=3840&q=75)
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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