Energy dissipated in a circuit

In summary, the discussion involved calculating the emf of a cell connected in series with a resistor, given the current and energy dissipated in both components. After finding the power dissipated in each component, the voltage drop across each can be determined using Ohm's law. Adding these voltage drops gives the total emf of the cell, which is the sum of IR and Ir. After solving for the individual voltages, the emf was found to be 2 V.
  • #1
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
 
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  • #2
SUSUSUSUSUSUSUSU said:
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...
 
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  • #3
berkeman said:
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??
 
  • #4
SUSUSUSUSUSUSUSU said:
Emf= IR+Ir = 3/4 + 5/4 =2

Is that it??
That's what I got as well. Good work! :smile:
 
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  • #5
berkeman said:
That's what I got as well. Good work! :smile:
Thanks a loooot!
 
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1. What is energy dissipation in a circuit?

Energy dissipation in a circuit refers to the conversion of electrical energy into other forms of energy, such as heat or light, as it flows through a circuit. This occurs due to the resistance of the circuit components, which causes some of the electrical energy to be lost as heat.

2. How is energy dissipated in a circuit?

Energy is dissipated in a circuit when a current flows through a resistor or other resistive component. The resistance in the circuit causes the electrons to collide with atoms, generating heat and reducing the amount of energy available for the circuit to do work.

3. How is energy dissipation measured in a circuit?

The energy dissipation in a circuit can be measured using a power meter, which measures the voltage and current in the circuit and calculates the power (P = VI). This power value represents the rate at which energy is being dissipated in the circuit.

4. How does energy dissipation affect circuit performance?

Energy dissipation can have a significant impact on circuit performance. As energy is lost as heat, it can cause components to overheat and potentially fail. It also reduces the efficiency of the circuit, as less energy is available for it to do useful work.

5. How can energy dissipation be reduced in a circuit?

In order to reduce energy dissipation in a circuit, components with lower resistance can be used, such as high-quality wires and low-resistance resistors. Additionally, using circuit designs that minimize the flow of current can also help reduce energy dissipation. Regular maintenance and proper cooling systems can also help prevent overheating and reduce energy loss in a circuit.

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