KCET · Physics · Current Electricity
A galvanometer of resistance \( 50 \Omega \) gives a full scale deflection for a current \( 5 \times 10^{-4} \mathrm{~A} \). The resistance that should be connected in series with the galvanometer to read \( 3 \mathrm{~V} \) is
- A \( 595 \Omega \)
- B \( 5050 \Omega \)
- C \( 5059 \Omega \)
- D \( 5950 \Omega \)
Answer & Solution
Correct Answer
(D) \( 5950 \Omega \)
Step-by-step Solution
Detailed explanation
Resistance of galvanometer \(=50 \Omega\) current \(=5 \times 10^{-4} A\); voltage \(=3 \mathrm{~V}\)
We know resistance,
\(R=\frac{V}{I_{g}}-G=\frac{3}{5 \times 10^{-4}}-50=6000-50\)
\(\Rightarrow R=5950\)
Thus, resistance that should be connected in series with the galvanometer to read \(3 \mathrm{~V}\) is \(5950 \Omega\).
We know resistance,
\(R=\frac{V}{I_{g}}-G=\frac{3}{5 \times 10^{-4}}-50=6000-50\)
\(\Rightarrow R=5950\)
Thus, resistance that should be connected in series with the galvanometer to read \(3 \mathrm{~V}\) is \(5950 \Omega\).
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