COMEDK · Physics · 19. Current Electricity
A galvanometer having a resistance of \(8 \Omega\) is shunted by a wire of resistance \(2 \Omega\). If the total current is \(1 \mathrm{~A}\), the part of it passing through the shunt will be
- A 0.5 A
- B 0.25 A
- C 0.2 A
- D 0.8 A
Answer & Solution
Correct Answer
(D) 0.8 A
Step-by-step Solution
Detailed explanation
Let \(G\) be the resistance of the galvanometer and \(S\) be the resistance of the shunt. Given \(G = 8 \Omega\) and \(S = 2 \Omega\). The total current \(I = 1 \text{ A}\) divides into current through the galvanometer \(I_g\) and current through the shunt \(I_s\). The current…
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