990*R/(990+R) = 990/10.
990*R/(990+R) = 99.
990*R = (990+R)99.
10R = 990+R.
9R = 990.
R = 110 Ohms.
The sensitivity of a galvanometer of resistance 990ohm is increased by 10 times.The shunt used is??
3 answers
Note: Since the current is 10 times the
previous current, the resistance is 1/10
of the previous resistance: 990/10.
previous current, the resistance is 1/10
of the previous resistance: 990/10.
there is a small correction in the problem. the sensitivity does not increase but decreases.
the shunt is connected parallel to galvanometer. Imagine a circuit where main current is i and shunt is connected parallel to galvanometer. main current flowing is i into this combination. it is given that the sensitivity decreases by 10 times, the current flowing through the galvanometer become i / 10, the remaining current flows though the shunt that is 9 i/10. since shunt and galvanometer are in parallel the potential difference across both will be same, so
potential across galvanometer = potential across shunt
( using ohm's law)
(9 i/10) x s = (i / 10) x G
substitute the values given for G, so shunt s = 110 ohm.
the shunt is connected parallel to galvanometer. Imagine a circuit where main current is i and shunt is connected parallel to galvanometer. main current flowing is i into this combination. it is given that the sensitivity decreases by 10 times, the current flowing through the galvanometer become i / 10, the remaining current flows though the shunt that is 9 i/10. since shunt and galvanometer are in parallel the potential difference across both will be same, so
potential across galvanometer = potential across shunt
( using ohm's law)
(9 i/10) x s = (i / 10) x G
substitute the values given for G, so shunt s = 110 ohm.
1 answer