Electrical Engineering - Parallel Circuits - Discussion

Discussion Forum : Parallel Circuits - General Questions (Q.No. 1)
1.
An ammeter has an internal resistance of 50 . The meter movement itself can handle up to 1 mA. If 10 mA is applied to the meter, the shunt resistor, RSH1, is approximately
55
5.5
50
9
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
16 comments Page 2 of 2.

NITIN said:   1 decade ago
Res=RM/(m-1)
Where as RM=Meter Resistance
m=Multi factor(10mA/1mA=10)
Res=50/(10-1)=5.5ohm Ans This one.

Raviraj said:   1 decade ago
See is the meter resistance is ohm.it Carries max of 1mA current.Nowif wb Wanne mesure 10mA With Same meter then we have to connect one low resistance across it so that we can Bypass axtra 9mA current through it

So Formula is Res=RM/(m-1)
Where as RM=Meter Resistance
m=Multi factor(10mA/1mA=10)
Res=50/(10-1)=5.5ohm Ans This one.

Muna said:   1 decade ago
The resistance is calculated by voltage equtation.

R = 50*1/1000*1000/9 = 5.5ohm.

Dushyant Dutta said:   1 decade ago
Rg*Ig/I-Ig=S

Bharamu said:   1 decade ago
Hi Anand, here is the answer.

See the meter resistance is 50 ohm, it carries max of 1mA current. Now if we wanna measure 10mA with same meter then we have to connect one low resistance across it so that we can bypass extra 9mA current through it.

So formula is
Rsh = Rm/(m-1)
Where as Rm = Meter Resistance
m = Multiplying Factor ( 10mA/1mA = 10)
Rsh = 50/(10-1) = 5.5 ohm

Anand said:   1 decade ago
Can anybody explain the answer ?


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