A cell of emf E and internal resistance r is connected to two external resistance $R_{1}$ and $R_{2}$ and a perfect ammeter. The current in the circuit is measured in four different situation :
(i) Without any external resistance circuit
(ii) With resistance $R_{1}$ only
(iii) With $R_{1}$ and $R_{2}$ in series combination
(iv) With $R_{1}$ and $R_{2}$ in parallel combination
The current measured in the four cases are 0.42 A, 1.05 A, 1.4 A and 4.2 A, but not necessarily in that order. Identify the current corresponding to the four cases mentioned above

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Solution

Given: A cell of emf E and internal resistance r is connected to two external resistance $R_{1}$ and $R_{2}$ and a perfect ammeter. The current in the circuit is measured in four different situation:
(i) Without any external resistance circuit
(ii) With resistance $R_{1}$ only
(iii) With $R_{1}$ and $R_{2}$ in series combination
(iv) With $R_{1}$ and $R_{2}$ in parallel combination
To find the current corresponding to to the four cases
Solution:
The current in the circuit to corresponding situations is given by:
(i) When here is no external resistance in the circuit
$I_{1} = \frac{E}{r}$
The current in this case will be maximum because effective resistance is minimum. So $I_{1} = 4.2 A$
(ii) In the presence of resistance $R_{1}$ only, we have
$I_{2} = \frac{E}{r + R_{1}}$
Here, the effective resistance is more than (i) and (iv) but less than (iii).
So $I_{2} = 1.05 A$
(iii) When $R_{1}$ and $R_{2}$ are in series combination, we have
$I_{3} = \frac{E}{r + (R_{1} + R_{2})}$
In this case, effective resistance is maximum, so current is minimum.
Thus, $I_{3} = 0.42 A$
(iv) When $R_{1}$ and $R_{2}$ are in parallel combination, we have
$I_{4} = \frac{E}{r + (\frac{R_{1} R_{2}}{R_{1} + R_{2}})}$
Here, effective resistance is more than (i) but less than (ii) and (iii).
So $I_{4} = 1.4 A$

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