Electrical Engineering - Parallel Circuits - Discussion

Discussion :: 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

[A]. 55
[B]. 5.5
[C]. 50
[D]. 9

Answer: Option B

Explanation:

No answer description available for this question.

Anand said: (Jul 26, 2011)  
Can anybody explain the answer ?

Bharamu said: (Jul 27, 2011)  
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

Dushyant Dutta said: (Sep 16, 2011)  
Rg*Ig/I-Ig=S

Muna said: (Sep 17, 2011)  
The resistance is calculated by voltage equtation.

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

Raviraj said: (Jan 1, 2012)  
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.

Nitin said: (Jun 9, 2012)  
Res=RM/(m-1)
Where as RM=Meter Resistance
m=Multi factor(10mA/1mA=10)
Res=50/(10-1)=5.5ohm Ans This one.

Rohit said: (Jul 14, 2012)  
This is parallel circuit question. When current flow 10A but resistance is not vary so that parallel resistor is 50. Because 50*50/ (50+50).

Mohammed Irshad said: (May 15, 2013)  
As per the question, the circuit is:
Supplying current = Is = 10mA.
Ammeter current = Ia = 1mA.

The circuit is in parallel, so it is current divider circuit.
Therefore,

Ish = Is-Ia = 10mA-1mA = 9mA.

Va = Vsh = Vs = Ra*Ia = 50*1*10^-3 = 50mV.

Rsh = Vsh/Ish = (50*10^-3)/(9*10^-3) = 5.55V.

Answer B is correct.

Sumant said: (Sep 8, 2013)  
Supplying current = Is = 10mA.
Ammeter current = Ia = 1mA.

The circuit is in parallel, So it is current divider circuit.

Therefore,

Ish = Is-Ia = 10mA-1mA = 9mA.

Va = Vsh = Vs = Ra*Ia = 50*1*10^-3 = 50mV.

Rsh = Vsh/Ish = (50*10^-3)/(9*10^-3) = 5.55V.

Answer B is correct.

Sharan said: (Aug 19, 2017)  
Well explained @Sumant.

David said: (Jan 30, 2018)  
Good explanation @Sumant.

Bibek said: (Apr 16, 2018)  
Rsh=(Im*Rm)/(I-Im).
Im=1mA,Rm=50, I=10mA.

Put the value in above eqn.
we will get;
Rsh=5.5.

Harris said: (Nov 21, 2018)  
There is a much simpler formula to solve this.

R(sh)= Ig/(It-ig) * Rg.

Ig= current through meter,
It= Total Current,
Rg= Meter Resistance,

And we get; (0.001/0.009) * 50 = 5.5 ohms.

Hemanth said: (Jun 21, 2019)  
The resistance is calculated by voltage equtation.

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

Arjun said: (Aug 14, 2019)  
Rsh = Rm/(m-1).
Where as Rm = Meter Resistance.
m = Multiplying Factor ( 10mA/1mA = 10).
Rsh = 50/(10-1) = 5.5 ohm.

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