Two identical moving coil galvanometer have 10Ω resistance and full scale deflection at 2μA current. One of them is converted into a voltmeter of 100 mV full scale reading and the other into an Ammeter of 1 mA full scale current using appropriate resistors. These are then used to measure the voltage and current in the Ohm's law experiment with R=1000Ω resistor by using an ideal cell. Which of the following statement(s) is/are correct ?
(A)The measured value of R will be 978Ω<R<982Ω.✓ Correct
(B)The resistance of the Voltmeter will be 100kΩ.
(C)The resistance of the Ammeter will be 0.02Ω (round off to 2nd decimal place)✓ Correct
(D)If the ideal cell is replaced by a cell having internal resistance of 5Ω then the measured value of R will be more than 1000Ω.
Answer: (A), (C)
Solution
Voltmeter conversion:
0.1=2×10−6(10+Rv)
Rv=49990Ω≈50 kΩ
Option (B) is incorrect (states 100 kΩ).
Ammeter conversion:
2×10−6×10=10−3RA
RA=0.02Ω
Option (C) is correct.
Measurement with ideal cell: Total voltmeter resistance = 50000 Ω Ammeter resistance = 0.02 Ω
The circuit has voltmeter (50000 Ω) in parallel with 1000 Ω resistor, and ammeter (0.02 Ω) in series.
Using voltage divider and current measurement:
Measured R≈980Ω
Option (A) is correct (978 Ω < R < 982 Ω).
Option (D): With internal resistance, the measured value would be less than 1000 Ω, not more. Incorrect.
JEE Advanced · 2019 · 27 May 2019 · Paper 1 · One or more correct options
