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A cell has an emf of 1.5 V and an internal resistance of 0.65 Q.
The cell is connected to a resistor R.
[0[8].[4] State what is meant by an emf of 1.5 V.
[2 marks]
I
[0] 8]-[2] The current in the circuit is 0.31 A.
Show that the total power output of the cell is approximately 0.47 W.
[1 mark]
[o[s).[3] Calcutate the energy dissipated per second in resistor R.
[2 marks]
energy dissipated per second = Ist
[o[si.[4] The cell stores 14 kJ of energy when it is fully charged. The cell's emf and internal resistance are constant as the cell f discharged.
Calculate the maximum time during which the fully-charged cell can deliver energy to
resistor R.
[2 marks)
maximum time = 5

A cell has an emf of 1.5 V and an internal resistance of 0.65 Q. The cell is connected to a resistor R. [0[8].[4] State what is meant by an emf of 1.5 V. [2 mar class=
A cell has an emf of 1.5 V and an internal resistance of 0.65 Q. The cell is connected to a resistor R. [0[8].[4] State what is meant by an emf of 1.5 V. [2 mar class=

Answer :

  • 1. A cell has an emf of 1.5 V means  a battery with an Emf rating of 1.5v can supply a maximum of 1.5v of electrical energy in an open circuit.
  • Once it is connected to a closed circuit, it will have a potential difference of 1.5v(considering it has negligible resistance).

2.     Voltage(V) in the circuit is 1.5 V and The current (I) in the circuit is 0.31 A.

So we have to find out power (p) in the circuit.

So  P = VI

       P= 1.5(0.31)

       P = 0.465

        P= 0.47 W

the total power output of the cell is approximately 0.47 W.

3.   We have to obtain  the energy dissipated per second in resistor R.

P = 0.465 - I²R

P = 0.465 - 0.31²(0.65)

P  = 0.402535 J/s

P =  0.40 J

Total 0.40 Joule per second energy has been dissipated from the resistor R.

4. The cell stores 14 kJ of energy when it is fully charged. The cell's emf and internal resistance are constant as the cell f discharged.

The energy dissipated per second(e) in resistor R is .40

Cell have total (E) 14 kJ (14000 J) of energy.

We have to calculate the maximum time (t) during which the fully-charged cell can deliver energy to resistor R.

[tex]\(E=\into{{e}_}}{t}\\\)[/tex]

14000 = 0.40 × t

t =14000  / 0.40  

t =  35 000 s or about 9.75 hrs.

Fully-charged cell can deliver energy to resistor R up to 9.75 hrs.

What is Electromotive Force?

Electromotive force, abbreviation E or emf, energy per unit electric charge that is imparted by an energy source, such as an electric generator or a battery. Energy is converted from one form to another in the generator or battery as the device does work on the electric charge being transferred within itself.

Learn more about Electromotive force:

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