A) The reading will be reduced to one-half.
B) The reading will not be affected.
C) The reading will be doubled.
D) The reading will be increased four-fold.
Correct Answer: C
Solution :
Let, resistance of bulb P and Q be R. So reading of ammeter before connecting Q to the circuit is \[I=\frac{V}{R}\] ?.(i) After connecting Q to the circuit, resultant resistance of the circuit is \[\frac{1}{{{R}_{eq}}}=\frac{1}{R}+\frac{1}{R}=\frac{2}{R}\]or \[{{R}_{eq}}=\frac{R}{2}\Omega \] Thus, new reading of ammeter will be \[l'=\frac{V}{R/2}=\frac{2V}{R}\] or \[l'=2l\] (From (i))
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