AP EAMCET · PHYSICS · Current Electricity
A galvanometer of resistance \(40 \Omega\) gives a deflection of 10 divisions per \(\mathrm{mA}\). There are 50 divisions on the scale. Maximum current that can pass through it when a shunt resistance of \(2 \Omega\) is connected is
- A \(105 \mathrm{~mA}\)
- B \(155 \mathrm{~mA}\)
- C \(210 \mathrm{~mA}\)
- D \(75 \mathrm{~mA}\)
Answer & Solution
Correct Answer
(A) \(105 \mathrm{~mA}\)
Step-by-step Solution
Detailed explanation
Given, galvanometer resistance, \(R_G=40 \Omega\) Shunt resistance, \(R_\lambda=2 \Omega\) Reading \(=10 \mathrm{div} / \mathrm{mA}\) and number of divisions, \(n=50\) \(\therefore\) Galvanometer current, \(I_G=\frac{50}{10}=5 \mathrm{~mA}\) Let shunt current be \(I\).…
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