The length of potentiometer wire is 1200cm, and it carries a current of 80mA, for a cell emf of 4V and internal resistance of 20 ohm, null point is found to be at 100cm, if a voltmeter is connected across the cell, the balancing point is decreased by 20cm, find:-
1. The reading of voltmeter
2. The resistance of whole wire
3. Resistance of voltmeter.
Asked by sakshi65341 | 15th May, 2019, 11:44: PM
When voltmeter is connected across the cell, we can consider resistance R of voltmeter is connected across the cell.
When a resistance R is connected across the cell, for potentiometer, we have (R+r)/R = l1 / l2 ...........(1)
where r is internal resistance of cell, l1 is the null point when voltmeter is not connected across cell
and l2 is the null point when voltmeter is connected across cell.
Hence substituting the values in eqn.(1), we get (R+20)/R = 100/80 ...............(2)
Hence from eqn.(2), we get Voltmeter resistance R = 80 Ω .
Reading of voltmeter :-
when voltmeter is connected across the cell, internal resistance and resistance of voltmeter are in series
and this series combination is connected across cell.
Hence current drawn by voltmeter = 4 V/(80+20)Ω = 40 mA .
Since reading of voltmeter is the potential difference across voltmeter resistance,
then we have voltmeter reading = 40×10-3×80 = 3.2 V
Resistance of whole wire :-
whenn cell alone is connected to potentiometer to get null point l1 , we have E = Φl1 ..................(3)
where E is the emf of cell, Φ is potential difference across unit length of wire and l1 is the null point length.
By substituting the values in eqn.(3), we have 4 = Φ×100 or Φ = 4/100 V/cm .
Hence the potential difference across full length of wire = (4/100)×1200 = 48 V.
Since the wire carries 80 mA current, then resistance of full length of wire = 48V/80 mA = 600 Ω
Answered by Thiyagarajan K | 16th May, 2019, 02:56: PM
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