Meter Circuit: Calculating Scale Reading with and without Additional Resistor

[thereddevils]

Homework Statement

Meter in the circuit shown has a linear scale that has not been calibrated and the scale reading is 20 . When another resistor of resistance 2000 ohm is connected across XY . What is the scale reading of the meter

Homework Equations

The Attempt at a Solution

Without the 2000 ohm resistor , I= 1.5 /(2995 + 50) , I = 0.0005

the reading given by the ammeter is 40600 times of the actual current .

now with the resistor , I = 1.5/(2995 + 50+2000) , I = 0.0003

the reading by the ammater will be 0.0003 x 40600 = 12

but the answer given is 25

 

[rl.bhat]

Without the 2000 ohm resistor , I= 1.5 /(2995 + 50) , I = 0.0005
This is wrong. In the given circuit the total resistance is (2995+50 +2000)
Find the current I.
I = k*θ. where k is the constant of the meter and θ is the deflection in the mete.
When you connect another 2000 ohm resistance in parallel with 2000 ohm resistance in the circuit, what is net resistance in the circuit? Find the current and equate it to k*θ'. find θ'.

 

[thereddevils]

Without the 2000 ohm resistor , I= 1.5 /(2995 + 50) , I = 0.0005
This is wrong. In the given circuit the total resistance is (2995+50 +2000)
Find the current I.
I = k*θ. where k is the constant of the meter and θ is the deflection in the mete.
When you connect another 2000 ohm resistance in parallel with 2000 ohm resistance in the circuit, what is net resistance in the circuit? Find the current and equate it to k*θ'. find θ'.

got it ! Thanks !