COMEDK · Physics · 19. Current Electricity
The resistance of the galvanometer and shunt of an ammeter are \(90 \mathrm{~ohm}\) and \(10 \mathrm{~ohm}\) respectively, then the fraction of the main current passing through the galvanometer and the shut respectively are:
- A \(\dfrac{1}{10}\) and \(\dfrac{1}{90}\)
- B \(\dfrac{1}{10}\) and \(\dfrac{9}{10}\)
- C \(\dfrac{9}{10}\) and \(\dfrac{1}{10}\)
- D \(\dfrac{1}{90}\) and \(\dfrac{1}{10}\)
Answer & Solution
Correct Answer
(B) \(\dfrac{1}{10}\) and \(\dfrac{9}{10}\)
Step-by-step Solution
Detailed explanation
Let \(I\) be the total current, \(I_g\) be the current through the galvanometer, and \(I_s\) be the current through the shunt. Given the resistance of the galvanometer \(G = 90 \, \Omega\) and the resistance of the shunt \(S = 10 \, \Omega\). Since the galvanometer and shunt are…
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