Work done by an electric cell in moving a charge of 10C is 5J. What is the EMF of the cell?
Answers
Answered by
1
Answer is B.
The electromotive force (e) or e.m.f. is the energy provided by a cell or battery per coulomb of charge passing through it, it is measured in volts (V). It is equal to the potential difference across the terminals of the cell when no current is flowing.
e = E/Q
e = electromotive force in volts, VE = energy in joules, JQ = charge in coulombs, C
In this case, work done = 5 J and amount of charge = 10 C.
Therefore, e = 5/10 = 1/2.
Hence, electromotive force of the cell is 1/2 volt.
The electromotive force (e) or e.m.f. is the energy provided by a cell or battery per coulomb of charge passing through it, it is measured in volts (V). It is equal to the potential difference across the terminals of the cell when no current is flowing.
e = E/Q
e = electromotive force in volts, VE = energy in joules, JQ = charge in coulombs, C
In this case, work done = 5 J and amount of charge = 10 C.
Therefore, e = 5/10 = 1/2.
Hence, electromotive force of the cell is 1/2 volt.
Answered by
4
Answer:
0.5 Volts is the E.M.F of the cell.
Explanation:
Given :-
Charge (q) = 10 c
Work done (W) = 5 J
To find :-
E.M.F of the cell (Electromotive force)
Solution :-
➤ E.M.F of a cell (E) = W/q
⇒ 5/10
⇒ 1/2
⇒ 0.5
∴ The E.M.F of the cell is 0.5 V.
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