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.
Mq.Khan said:
4 years ago
Both current and resistance is known for 50 ohms.
Hence;
Voltage = 50 x 1 mA = 0.05.
Voltage is same in parallel.
Hence for unkown resistor
R= V/I.
R= 0.05/9 mA = 5.5 ohm.
Hence;
Voltage = 50 x 1 mA = 0.05.
Voltage is same in parallel.
Hence for unkown resistor
R= V/I.
R= 0.05/9 mA = 5.5 ohm.
(6)
Arjun said:
6 years ago
Rsh = Rm/(m-1).
Where as Rm = Meter Resistance.
m = Multiplying Factor ( 10mA/1mA = 10).
Rsh = 50/(10-1) = 5.5 ohm.
Where as Rm = Meter Resistance.
m = Multiplying Factor ( 10mA/1mA = 10).
Rsh = 50/(10-1) = 5.5 ohm.
(10)
Hemanth said:
6 years ago
The resistance is calculated by voltage equtation.
R = 50*1/1000*1000/9 = 5.5ohm.
R = 50*1/1000*1000/9 = 5.5ohm.
(2)
Harris said:
7 years ago
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.
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.
(2)
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.
David said:
8 years ago
Good explanation @Sumant.
Sharan said:
8 years ago
Well explained @Sumant.
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)
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.
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).
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