Four resistance of 10 ,60 , 100 and 200 respectively taken in order are used to form a Wheatstone’s bridge.A 15V battery is connected to the ends of a 200 resistance, the current through it will be
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Here three reistance for 10ohms ,60ohm and 100ohm are in series and they are parallel to 200ohm reisitance .
When a potenital difference of 15v applied to 200ohm, then current I=V/R
I=15/200
=7.5 X 10 ⁻² A
When a potenital difference of 15v applied to 200ohm, then current I=V/R
I=15/200
=7.5 X 10 ⁻² A
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R1=10 ohm
R2=60 ohm
R3=100 ohm
R4=200 ohm
Condition for a wheatstone's bridge
R2/R1=R3/R4
the ratio of resistances will not be equal for the given data
therefore it will not form a wheatstone's bridge
R2=60 ohm
R3=100 ohm
R4=200 ohm
Condition for a wheatstone's bridge
R2/R1=R3/R4
the ratio of resistances will not be equal for the given data
therefore it will not form a wheatstone's bridge
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