A) ratio of potential differences across them is 3/2
B) ratio of potential differences across them is 9/4
C) ratio of powers consumed across them is 4/9
D) ratio of powers consumed across them is 2/3
Correct Answer: C
Solution :
The power of bulb \[P=\frac{{{V}^{2}}}{R}\] \[R=\frac{{{V}^{2}}}{P}\] \[\therefore \] \[\frac{{{R}_{1}}}{{{R}_{2}}}={{\left( \frac{200}{300} \right)}^{2}}=\frac{4}{9}\] When connected in series potential drop and power consumed are in the ratio of their resistances. So, \[\frac{{{P}_{1}}}{{{P}_{2}}}=\frac{{{V}_{1}}}{{{V}_{2}}}=\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{4}{9}\]
You need to login to perform this action.
You will be redirected in
3 sec
