Question : A cell of emf E and internal resistance r is connected to two external resistances R1 and R2 and a perfect ammeter. The current in the circuit is measured in four different situations:
(i) without any external resistance in the circuit
(ii) with resistance R1 only
(iii) with R1 and R2 in series combination
(iv) with R1 and R2 in parallel combination
The currents 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 currents corresponding to the four cases
mentioned above.
Doubt by Lakshita
Solution :
Here the ammerter used is perfect so we can say that it has negligible internal resistance.
Hence, the amount of current in circuit will be given by
I = E/(R+r)
Where
E = Emf of the cell
Hence, the amount of current in circuit will be given by
I = E/(R+r)
Where
E = Emf of the cell
External R = Resistance of the cell
r = Internal Resistance of the cell
r = Internal Resistance of the cell
Case (i)
Here the value of external resistance is zero i.e. R=0
So equivalent resistance of the circuit will be R(i) = r
Case (ii)
Here the value of external resistance is R1 i.e. R=R1
So equivalent resistance of the circuit will be R(ii) = r+R1
Case (iii)
Here R1 and R2 are connected in series so total external resistance will be (R1+R2)
So equivalent resistance of the circuit will be R(iii) = r+(R1+R2)
Case (iv)
Here R1 and R2 are connected in parallel so total external resistance will be R1R2/(R1+R2)
So equivalent resistance of the circuit will be R(iv) = r+[R1R2/(R1+R2)]
Clearly,
R(iii)>R(ii)>R(iv)>R(i)
R(iii)>R(ii)>R(iv)>R(i)
Now
I = E/Req
E = constant
I = E/Req
E = constant
I∝1/Req
Hence,
I(iii)<I(ii)<I(iv)<I(i)
Hence,
I(iii)<I(ii)<I(iv)<I(i)
0.42 A<1.05 A<1.4A<4.2A
So,
In case (i) I = 4.2 A
In case (ii) I = 1.05 A
In case (iii) I = 0.42 A
In case (iv) I = 1.4 A
In case (iv) I = 1.4 A
