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 1 of 2.
Anand said:
1 decade ago
Can anybody explain the answer ?
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
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:
1 decade ago
Rg*Ig/I-Ig=S
Muna said:
1 decade ago
The resistance is calculated by voltage equtation.
R = 50*1/1000*1000/9 = 5.5ohm.
R = 50*1/1000*1000/9 = 5.5ohm.
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.
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:
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.
Sumant said:
1 decade ago
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.
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.
(2)
Sharan said:
8 years ago
Well explained @Sumant.
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