Question

1. An ammeter of resistance 2012 gives full scale
deflection, when 1 mA current flows through it.
What is the maximum current that can be
measured by connecting 4 resistors each of
1612 in parallel with the ammeter?
(a) 6 mA (b) 4 mA (c) 8 mA (d) 2 mA​

Answer 1

Answer:

OptionA)6mA

Explanation:

Given, 

Meter has a resistance G = 20 Ω

Current through meter for full deflections 

tg=1mA=1×10−3A

Now, for stunt three resistors each have resistance 12 Ω are connected in parallel.

Hence, 

Rp1=121+121+121=123⇒Rp=4Ω

Now, from the formula

R=I−IgIg ...(i)

( where, R is shunt resistance R = 4 Ω )

Now, putting the values in Eq. (i)

4=I−1×10−31×10−3×20

4I−4×10−3=20×10−3

4I=20×10−3+4×10−3=24×10−3

I=424×10−3=6×10