A cell of unknown EMF E and internal resistance r, two unknown resistances R1 and R2 (R2>R1) and a perfect ammeter are given. The current in the circuit is measured in five different situations :
- without any external resistance in the circuit,
- with resistance R1 only,
- with resistance R2 only,
- with both R1 and R2 used in series combination, and
- with both R1 and R2 used in parallel combination.
The current obtained in the five cases are 0.42 A, 0.6 A, 1.05 A, 1.4 A and 4.2 A, but not necessarily in that order.
In terms of r, what is the value of resistance R2 ?
Answer & explanation
Correct answer: option 3
The current will be maximum when the resistance is in the lowest value configuration.
$ \frac{E}{r} = 4.2$ .........(1)
$\frac{E}{r+R_1} = 1.05 A$ .......(2)
$\frac{E}{r+R_2} = 0.6 A$ ...........(3)
$\frac{E}{r+R_1+R_2} = 0.42A$ ........(4)
$\frac{E}{r + \frac{R_1 R_2}{R_1+R_2}} = 1.4A$ ...........(5)
From Equation (1) ,$ E = 4.2r$
$ \text{From equation (2) } , \Rightarrow \frac{4.2r}{r+R_1} = 1.05$
$\Rightarrow r+R_1 = 4r , R_1 = 3r$
$\text{ From equation 1 and 3 } \frac{4.2r}{r+R_2} = 0.6 $
$\Rightarrow 4.2 r = 0.6 r + 0.6 R_2 , R_2 = \frac{3.6}{0.6} = 6\Omega$