A galvanometer of resistance 0.80ohm can measure current 1A.Calculate the current value of resistance required so that it can measure upto 5A
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Explanation:
(i) The net resistance of ammeter with shunt connected in parallel=R
2
=
R
A
+x
R
A
x
=
0.8+x
0.8×x
The potential of the cell is
V=I
1
R
A
=I
2
R
2
⟹1×0.8=5×
0.8+x
0.8x
⟹x=0.2Ω
(ii) Combined resistance of the ammeter and thus shunt
R=
0.8+x
0.8x
=0.16Ω
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