What Is the EMF of a Cell with Given Energy Dissipation and Current?

[SUSUSUSUSUSUSUSU]

Homework Statement

A cell is connected in series with a resistor and supplies a current of 4.0 A for a time of 500 s. During this time, 1.5 kJ of energy is dissipated in the cell and 2.5 kJ of energy is dissipated in the resistor.

What is the emf of the cell?

A. 0.50 V

B. 0.75 V

C. 1.5 V

D. 2.0 V

cheers

Homework Equations

V=IR
P=IV=I^2R
P= energy/time
Emf= V+Ir

The Attempt at a Solution

Power dissipated in the cell is 1500 x 500 = 750000 W
Voltage of the cell= 750000/4 = 187500 V
Power dissipated in the resistor is 2500 x 500 = 1250000 W
Voltage of the cell= 1250000/4= 312500 V

However, I do not know how to refer those to the emf equation.
Emf= IR+Ir = 4(R+r)

I cannot go further than this

Could you guys please solve this question for me...?

Thanks

 

[berkeman]

Power dissipated in the cell is 1500 x 500 = 750000 W

Power is Energy divided by time, not multiplied by it. Does that help?

Basically use the Energy and time to find the power in Ri and Rl, then use the series current to figure out the voltage drop across each. Add those to get the total EMF of the battery...

 

[SUSUSUSUSUSUSUSU]

Power is Energy divided by time, not multiplied by it. Does that help?

Basically use the Energy and time to find the power in Ri and Rl, then use the series current to figure out the voltage drop across each. Add those to get the total EMF of the battery...

Thank you, you are right!

I got 3W for the cell and 5W for the resistor!

3=4xV.
Therfore V=3/4
5=4xV
Therefore V=5/4

Emf= IR+Ir = 3/4 + 5/4 =2

Is that it??

 

[berkeman]

Emf= IR+Ir = 3/4 + 5/4 =2

Is that it??

That's what I got as well. Good work!

 

[SUSUSUSUSUSUSUSU]

That's what I got as well. Good work!

Thanks a loooot!