If one more resistance of 50 ohm is connected in series what happens to the heat generated.
Answers
Answered by
0
Answer:
Current in the circuit without ammeter is I
max
=V/R=5/50=0.1A
Given that current in the circuit with ammeter is within 1% of original current.
Hence, current flowing is I
′
=0.99I
max
=0.099A
Resistance r
s
is connected in parallel with the Galvanometer to increase its range.
50Ω is connected in series to ammeter.Therefore, overall resistance of the circuit is:
R=50+
100+r
s
100r
s
By Ohm's law:
V=I
′
R
5 = 0.099× (50+100r
s
)/(100+r
s
)
⟹ r
s
= 0.5Ω
Answered by
0
Answer:
very heat
Explanation:
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