KCET · Physics · Alternating Current
What is the value of shunt resistance required to convert a galvanometer of resistance \( 100 \Omega \)
into an ammeter of range \( 1 \mathrm{~A} \)?
Given : Full scale deflection of the galvanometer is \( 5 \mathrm{~mA} \).
- A \( \frac{5}{9.95} \Omega \)
- B \( \frac{9.95}{5} \Omega \)
- C \( 0.5 \Omega \)
- D \( 0.05 \Omega \)
Answer & Solution
Correct Answer
(A) \( \frac{5}{9.95} \Omega \)
Step-by-step Solution
Detailed explanation
Given, resistance of galvanometer \( =100 \Omega \); range of ammeter, \( I=1 \mathrm{~A} \); full scale deflection \( I_{g}=5 \mathrm{~mA}=5 \times 10^{-3} A \)
Value of shunt, \( S=\frac{I_{g} G}{I-I_{g}} \)
Substituting the values, we get
\(S=\frac{5 \times 10^{-3} \times 100}{1-5 \times 10^{-3}}=\frac{0.5}{1-0.005}=\frac{0.5}{0.995} \Omega=\frac{5}{9.95} \Omega\)
Therefore, value of shunt resistance is \( \frac{5}{9.95} \Omega \)
Value of shunt, \( S=\frac{I_{g} G}{I-I_{g}} \)
Substituting the values, we get
\(S=\frac{5 \times 10^{-3} \times 100}{1-5 \times 10^{-3}}=\frac{0.5}{1-0.005}=\frac{0.5}{0.995} \Omega=\frac{5}{9.95} \Omega\)
Therefore, value of shunt resistance is \( \frac{5}{9.95} \Omega \)
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