Two electric bulbs whose resistances are in the ratio of 1 : 2 are connected in parallel to a constant voltage source. The powers dissipated in them have the ratio.

a
1 : 2

b
1 : 1

c
2 : 1

d
1 : 4

Question

Two electric bulbs whose resistances are in the ratio of 1 : 2 are connected in parallel to a constant voltage source. The powers dissipated in them have the ratio.

a	
     1 : 2

b	
     1 : 1

c	
     2 : 1

d	
     1 : 4

anonymous · Answer

P = V^2/R
So, the smaller the resistance, the higher the power. Since you had 1 to 2 resistances, you now will have 2 to 1 power dissipations

anonymous · Answer

to be more clear about why it's 2:1, basically, imagine that P_1 is dissipated by R_1. To find P_2, simply replace R_2 by R_1 in the equation :
$$P_2 = V^2/R_2 = V^2/(2R_1) = 0.5 (V^2/R_1) = 0.5 P_1$$

anonymous · Answer

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