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
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
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