Electrical Engineering - Series-Parallel Circuits - Discussion
Discussion Forum : Series-Parallel Circuits - General Questions (Q.No. 11)
11.
The internal resistance of a 30,000 ohm/volt voltmeter set on its 50 V range is
15,000 
150,000
1,500,000
15,000,000
Discussion:
7 comments Page 1 of 1.
Mehr said:
6 years ago
R2/V2=R1/V1.
as 30000ohm/1v*50v.
R2=1500000ohm.
as 30000ohm/1v*50v.
R2=1500000ohm.
Ramesh said:
7 years ago
The voltmeter has an internal resistance of 30,000 ohm/volt.
That is for 1volt the resistance is 30,000 ohm.
Therefore,
For 50v the resistance.
50*30, 000 :-1,500,000Ω.
That is for 1volt the resistance is 30,000 ohm.
Therefore,
For 50v the resistance.
50*30, 000 :-1,500,000Ω.
Alex Smylez said:
8 years ago
IR=ohm/volt*voltage range.
which is 30000*50=1500000
which is 30000*50=1500000
Hems said:
1 decade ago
Can one explain clearly about this question?
(1)
Ninad sonawane said:
1 decade ago
Internal resistance = ohm/volt * vtg range.
So 30000*50 = 1500000.
So 30000*50 = 1500000.
Hema said:
1 decade ago
Given:
Internal Resistance = 30,000 ohm.
Voltage = 50V.
IR*V = 30,000*50 = 1500000.
Internal Resistance = 30,000 ohm.
Voltage = 50V.
IR*V = 30,000*50 = 1500000.
Anil said:
1 decade ago
30000*50=1500000(ans)
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