Electronics and Communication Engineering - Measurements and Instrumentation - Discussion
Discussion Forum : Measurements and Instrumentation - Section 3 (Q.No. 50)
50.
If a shunt of 200 Ω resistance is used with a galvanometer of 1000 Ω resistance, the multiplying power is
Answer: Option
Explanation:
If current through galvanometer is 1,
A current through shunt = 
. Total current = 6 I. Hence multiplying factor is 6.
Discussion:
5 comments Page 1 of 1.
Ece student said:
1 decade ago
The solution you have shown is not clear. Please explain.
Prem said:
9 years ago
Multiplying factor = I/Ic,
= (1 + Rc/Rs),
= (1 + 1000/200) = 6.
= (1 + Rc/Rs),
= (1 + 1000/200) = 6.
Rajashekar said:
8 years ago
Multiply factor=(Rm÷Rsh)+1.
Prashant Dhokiya said:
4 years ago
Multiplying power = {1+ (Rm÷Rsh) }.
Rm= galvanometer Internal Resistance,
Rsh= shunt resistance.
Rm= galvanometer Internal Resistance,
Rsh= shunt resistance.
MATIULLAH KAKAR said:
2 years ago
m = Shunt resistance + Galvanometer resistance/shunt resistance.
m =200 + 1000/200.
= 1200/200 = 6.
m =200 + 1000/200.
= 1200/200 = 6.
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