JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
In the circuit diagrams \((A, B, C\) and \(D\)) shown below, \(R\) is a high resistance and \(S\) is a resistance of the order of galvanometer resistance \(G\). The correct circuit, corresponding to the half deflection method for finding the resistance and figure of merit of the galvanometer, is the circuit labelled as
- A Circuit \(A\) with \(G = \frac{{RS}}{{R - S}}\)
- B Circuit \(B\) with \(G\, = S\)
- C Circuit \(C\) with \(G\, = S\)
- D Circuit \(D\) with \(G = \frac{{RS}}{{R - S}}\)
Answer & Solution
Correct Answer
(D) Circuit \(D\) with \(G = \frac{{RS}}{{R - S}}\)
Step-by-step Solution
Detailed explanation
The correct circuit diagram is \(D\) with galvanometer resistance \(G = \frac{{RS}}{{R - S}}\)
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