A) P/3
B) 3P
C) 9P
D) P/9
Correct Answer: C
Solution :
Key Idea: Power dissipated is inversely proportional to resistance. From Joules law, the power dissipated through a resistor R, having a potential difference V is: \[P=\frac{{{V}^{2}}}{R}\] When bulbs are connected in series \[R=R+R+R=3\,R\] Power dissipated is \[P=\frac{{{V}^{2}}}{3\,R}\] ?. (i) When they are connected in parallel \[\frac{1}{R}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}=\frac{3}{R}\] \[\Rightarrow \] \[R=\frac{R}{3}\]. Power dissipated is,\[P=\frac{{{V}^{2}}}{R/3}\] ... (ii) From Eqs. (i) and (ii), we have \[P=9\,P\].
You need to login to perform this action.
You will be redirected in
3 sec
