Example 27.7 Power in an Electric Heater Problem An electric heater is construct

Question

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 . Find the current carried by the wire and the power rating of the heater. Solution Because V = IR, we have the following. 1, 120 V R 7.00 We can find the power rating using the expression IR. 2 go = 12R = 12(7.00 ) = 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, g-(AV)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= le the ac tual povet consumption now (B) If we connect it a 240 V source, assuming that it still works, what will be the actual power consumption now (in kW)? will kW

Explanation / Answer

27.7) A) Power = V^2 / R

2000 = 120^2 / R

R = 7.2 ohm

V = IR

120 = I*7.2

I = 16.67 A

b) Po wer = 240^2 / 7.2 = 8000 W

= 8kW

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.