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A cell of emf E is connected across a resistance R. The potential difference across the terminals of the cell isfound to be V volt. Internal resistance of cell is a] (E-V)R b] (E-V)R/V c] 2(E-V)R/E d] 2(E-V)V/R

A cell of emf E is connected across a resistance R. The potential difference across the terminals of the cell isfound to be V volt. Internal resistance of cell is
a] (E-V)R
b] (E-V)R/V
c] 2(E-V)R/E
d] 2(E-V)V/R

Grade:12

2 Answers

Aditya Srivastava
24 Points
6 years ago
Assume a circuit (with the given conditions) :
Emf(E) , Internal Resistance(r), External Resistance(R)
When no current is drawn from the cell :
Emf(E) = Internal Potential Difference(V”) //across the plates of the emf.
But, when current(i) is drawn :
Emf(E) = V” + V //due to potential drop, external potential difference is obtained (V) ; as mentioned.
Now, since : V” = i . r ;
     Then, E = V” + V will become,
>  E = (i . r) + V
>  E – V = i . r
>  (E – V) / i = r
since, i = V / R ;
>  (E – V) / (V / R) = r
>  {(E – V) . R} / V = r  //ans(b)
ankit singh
askIITians Faculty 614 Points
3 years ago

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 = I(R+r)
E = electromotive force in volts, V
I = current in amperes, A
R = resistance of the load in the circuit in ohms,
r = internal resistance of the cell in ohms.
That is, E = IR+Ir, E = V + Ir.
Rearranging the equation we get, r=RE−V​.
Substituting I=V/R in the above equation, we get, r=RV​E−V​=V(E−V)R​.
Hence, internal resistance r=V(E−V)R​.

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