A galvanometer having 50 divisions provided with a variable shunt S used to measure the current as ammeter when connected in series with a resistance of 90 ohm and a battery of internal resistance 10 ohm, it is observed that when the shunt resistance are 10 and 50 ohm, the deflection are respectively 9 and 30 divisions. what is the resistance of the galvanometer? Further if the full scale deflection of the galvanometer movement is 300mA,find emf of the cell.
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shashankavsthi:
answer is 233.3ohms and 144V
Answered by
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Let the emf of the battery is E and resistance of galvanometer is r then,
According to question
.3×10^−3×9/50=E/90+10r/10+r−−−−(1)
And,
.3×10^−3×3/50=E/90+50r/50+r−−−(2)
Divide first by second
3=90+50r/50+r/90+10r/10+r
From here you will get value of r
r= −2.43 and −58.81 or
r=2.43,58.81ohm
So,
For these r there is two possible battery put the value of r in to equation 1 we getE=5mV an 1.8mV respectively for the resistances.
According to question
.3×10^−3×9/50=E/90+10r/10+r−−−−(1)
And,
.3×10^−3×3/50=E/90+50r/50+r−−−(2)
Divide first by second
3=90+50r/50+r/90+10r/10+r
From here you will get value of r
r= −2.43 and −58.81 or
r=2.43,58.81ohm
So,
For these r there is two possible battery put the value of r in to equation 1 we getE=5mV an 1.8mV respectively for the resistances.
r=233.3 ohm and emf =144
bcoz answwr is not matching that's why incorrect
his answer is also incorrect
u have any other answer?
koi nii then let others to do it bcoz i need its answer and i don't know how to do it.
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