Electrical Engineering - Series-Parallel Circuits - Discussion

Discussion Forum : Series-Parallel Circuits - General Questions (Q.No. 1)
1.
The internal resistance of a 20,000 ohm/volt voltmeter set on its 5 V range is
20,000
100,000
200,000
1,000,000
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
32 comments Page 2 of 4.

Junaid said:   9 years ago
Not understanding. Please explain me to get it.

Apoorva chourasiya said:   9 years ago
This is sensitivity concept.
Sensitivity = 2000ohm/volt.
Voltage = 5v
So, s = ohm/volt.

Ohm = s * volt.
Ohm = 2000 * 5,
Ohm = 10000ohm.

Nilesh Gitte said:   9 years ago
Correct explanation @M.Nedunchezhian.

Sanjay Meena said:   9 years ago
Om's I = V/R rule formula apply.

Om prakash said:   9 years ago
Very good @Juilee.

Sagar Kalekar said:   10 years ago
I = 1/(Ifsd).

Ifsd = Full scale deflection i.e. Sensitivity of the meter.

I = 1/20000 = 0.5 x 10^-4.

R = V/I.

R = 5/0.5 x 10^-4.

R = 100, 000 ohm.

Raji said:   1 decade ago
Don't think complicate manner just think about logic its simple the question is per volt it.

Consumes om 20000 there fore 5 volts just multiply it.

LALITHA said:   1 decade ago
Given,

r = 20,000 ohms per volt.
For 1v r = 20,000.
For 5v R = 20,000*5 = 100,000.

ZERO said:   1 decade ago
Don't tell me that IR is internal resistance? base on the ohms law I= amperes R= resistance, and E= volts. IR=V? in the ohms law pyramid not this IR*V?

Nazim uddin said:   1 decade ago
The internal resistant 20000 for 1 volt that = v = 5 and ohm law,
R = v*I 5*20000 = 100,000 ohm.


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