An ammeter with resistance `R_A` is connected in series with a resistor R, a battery of emf c and internal resistance `r`. The current measured by the ammeter is `I_A`. Find the current through the circuit if the ammeter is removed so that the battery and the resistor form a complete circuit. Express your answer in terms of `I_A, r, R_A` and `R`. Show that more "ideal" the ammeter, the smaller the difference between this current and the current `I_A`. ltbr. (b) If `R = 3.80 Omega, epsilon = 7.50 V` and `r = 0.45 Omega`, find the maximum value of the ammeter resistance `R_A` so that `I_A` is within `99%` of the current in the circuit when the ammeter is absent. (c) Explain why your answer in part (b) represents a maximum value.

Answer

A, B, D

Explanation

a. `I_A=epsilon/(R+R_A+r)`
`:. epsilon =(R+R_A+r)I_A`
Now, `I'_A=epsilon/(R+r)`……….i
Substituting the value of `epsilon` we get
`I'A=I_A(1+R_A/(R_A+r))`
If `R_Ararr 0.I_A^'gtI_A`
b. in eq. i substituiting
`I_A=0.99I'_A` and the given values we get
`R_A=0.0045 Omega`
`c. I_A=I_A^'/(1+R_A/(R_A+r))=I_A^'/(1+1/(1+R_A))`
If `R_A` is decreased from this value then `I_A` will increase from `99%` of `I'_A`