An ammeter gives full deflection when a current of 2 amp. flows throug...
Solution:
Given, full deflection current, I = 2 A and ammeter resistance, R = 12 Ω
Let the resistance to be connected in parallel with the ammeter be R'.
Let the maximum current that can be measured by the ammeter be Imax = 5 A.
To find: The value of R' to be connected in parallel with the ammeter.
Using the ammeter and R', a parallel combination is formed, which can be represented as shown below.
Let I' be the current that flows through the parallel combination.
From Kirchhoff's current law, the current I flowing through the circuit is given by:
I = I' + Imax
As per the question, the ammeter gives full deflection when the current is 2 A.
Therefore, the current through the ammeter, Ia = 2 A.
From Ohm's law, the potential difference across the ammeter, Va = Ia × R = 2 × 12 = 24 V.
The potential difference across the parallel combination, V' = Va.
From Ohm's law, the current I' flowing through the parallel combination is given by:
I' = V' / R'
Substituting the values of V' and I' in the equation obtained above, we get:
I = V' / R' + Imax
Substituting the values of I, V' and Imax in the above equation, we get:
5 = 24 / R' + 2
Solving the above equation for R', we get:
R' = 8 Ω
Therefore, the value of R' to be connected in parallel with the ammeter is 8 Ω.
Hence, option C is the correct answer.
An ammeter gives full deflection when a current of 2 amp. flows throug...