In the given circuit: The ratio between the reading of ammeter before and after closing the key is \( \qquad \) (A) \( \frac{10}{9} \) (B) \( \frac{5}{3} \) (C) \( \frac{9}{10} \) (D) \( \frac{3}{5} \)
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The resistors are in series, so the total resistance \( R_{\text{before}} \) is the sum of all resistors: \[ R_{\text{before}} = R + R + R = 3R \] Show more…
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