Electricity - Resistance problem

In summary, the problem asks for the resistance of a 240 W television that operates for 4.0 hours a day in a typical Canadian home. Using various equations, the power and energy values are given, but the current or voltage must be found in order to calculate the resistance. It's important to consider the relationship between power, voltage, current, and resistance, as well as the characteristics of electricity supplied to Canadian homes.
  • #1
gvhunting
1
0

Homework Statement



A 240 W television operates for 4.0 h a day in a typical Canadian home.
Calculate the resistance of the television.

Homework Equations



1) I = Q/∆ t 2) E = QxV 3) P = (Q x V)/ ∆t 4) P = I^2 x R
5) P = V^2/R 6) P = I x V 7) V = W/Q 8) V = E/Q
9) Q = I x ∆t 10) R = V/I


The Attempt at a Solution



Given:

P = 240 W
∆t = 4 h = 14 400 s
R = ?

E = P x ∆t
= 240 W x 14 400 s
= 3 456 000 J

I have done many practice problems but none such as this. I do not know where to go from here. There are many equations and I understand the most logical step next would be to either find the current or voltage and then using that value and Power to find the resistance. Or I can also find the value of Q and then using that and Power find either the current or voltage but not many values are given to do that.

Please help, I have been at this question for quite a while now, going around in circles.
 
Physics news on Phys.org
  • #3
gvhunting said:

Homework Statement



A 240 W television operates for 4.0 h a day in a typical Canadian home.
Calculate the resistance of the television.

Homework Equations



1) I = Q/∆ t 2) E = QxV 3) P = (Q x V)/ ∆t 4) P = I^2 x R
5) P = V^2/R 6) P = I x V 7) V = W/Q 8) V = E/Q
9) Q = I x ∆t 10) R = V/I


The Attempt at a Solution



Given:

P = 240 W
∆t = 4 h = 14 400 s
R = ?

E = P x ∆t
= 240 W x 14 400 s
= 3 456 000 J

I have done many practice problems but none such as this. I do not know where to go from here. There are many equations and I understand the most logical step next would be to either find the current or voltage and then using that value and Power to find the resistance. Or I can also find the value of Q and then using that and Power find either the current or voltage but not many values are given to do that.

Please help, I have been at this question for quite a while now, going around in circles.
Hi gvhunting, Welcome to Physics Forums.

How is power related to voltage and current? How about resistance? What are the common characteristics of the electricity supplied to homes in Canada?
 

Related to Electricity - Resistance problem

1. What is electrical resistance?

Electrical resistance is the measure of opposition to the flow of electrical current through a material. It is measured in ohms (Ω) and is influenced by factors such as the type of material, its length and thickness, and temperature.

2. How does resistance affect the flow of electricity?

Resistance causes a decrease in the flow of electrical current. The higher the resistance, the lower the current flow. This is described by Ohm's Law, which states that the current (I) is equal to the voltage (V) divided by the resistance (R).

3. What factors can increase resistance in a circuit?

Resistance can be increased by using materials with higher resistance, increasing the length of the material, or decreasing its thickness. Additionally, temperature can also affect resistance, with most materials having higher resistance at higher temperatures.

4. How can resistance be reduced?

Resistance can be reduced by using materials with lower resistance, shortening the length of the material, or increasing its thickness. Additionally, keeping the temperature low can also help to reduce resistance.

5. How does resistance affect the brightness of a light bulb?

Resistance plays a crucial role in determining the brightness of a light bulb. A higher resistance in the circuit will result in a lower current flow, which in turn will result in a dimmer light. On the other hand, a lower resistance will allow for a higher current flow, resulting in a brighter light.

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
843
  • Introductory Physics Homework Help
Replies
2
Views
845
  • Introductory Physics Homework Help
Replies
18
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
260
  • Introductory Physics Homework Help
Replies
12
Views
296
  • Introductory Physics Homework Help
Replies
1
Views
988
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
26
Views
1K
Replies
11
Views
491
Back
Top