Question

A real battery has an emf $\xi $ and an internal resistance r. A variable resistor R is connected across the terminal of the battery. A current I is drawn from the battery and the potential difference across the terminal on the battery is V. If R is slowly decreased to zero, which of the following best describes I and V?
A. I decreases to zero, V approaches
B. I approaches an infinite value, V decreases to zero
C. I approaches ${\xi }/{r}\;$ , V approaches
D. I approaches ${\xi }/{r}\;$, V decreases to zero

Answer 1

Hint: At first look out the values given in the question. Now we have to write the relation between terminal voltage V, E M F, current and resistance. In this relation we have to replace the value of V with ohm’s law. Now we can calculate that if R approaches zero then what will happen.


Complete step by step answer:
In the question it is given that,
A battery has an EMF $\xi $.
It has an internal resistance r.
The circuit has a Variable resistor R.
We know that the relation between V and EMF(E) is,
V = E – Ir.
Now, we are asked that if ‘R’ is reduced to zero then what is the effect on I and V.
Now in the above equation we do not see any ‘R’ so, we have to bring ‘R’ in the equation,
We know that, V = IR.
So, now we can write the equation as ,
IR = E – Ir.
Now, if we say that ‘R’ is slowly tending to zero then, ‘V’ is also tending to zero as V = IR.
So,
$0=\xi -Ir$,
$\xi =Ir$
$I=\dfrac{\xi }{r}$.
So in the first case we see that V is approaching towards zero as R is decreased to zero.
And we also see that $I$ is approaching $\dfrac{\xi }{r}$ as the ‘R’ is decreased to zero.
So, on the basis of the above explanation we can say that Option D is the correct option, that says I approach ${\xi }/{r}\;$, V decreases to zero.

Note:
In the equation V = E – Ir. ‘V’ is the potential drop across the load resistance. E is the EMF of the battery , I is the current passing through the battery and ‘r’ is the internal resistance in the circuit.