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
. The meter movement itself can handle up to 1 mA. If 10 mA is applied to the meter, the shunt resistor, RSH1, is approximately55
5.5
50
9
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.
Where as RM=Meter Resistance
m=Multi factor(10mA/1mA=10)
Res=50/(10-1)=5.5ohm Ans This one.
Rohit said:
1 decade ago
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:
1 decade ago
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.
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:
8 years ago
Well explained @Sumant.
David said:
8 years ago
Good explanation @Sumant.
Bibek said:
7 years ago
Rsh=(Im*Rm)/(I-Im).
Im=1mA,Rm=50, I=10mA.
Put the value in above eqn.
we will get;
Rsh=5.5.
Im=1mA,Rm=50, I=10mA.
Put the value in above eqn.
we will get;
Rsh=5.5.
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