Question about power with reference of a graph

[Clara Chung]

1. Homework Statement
As shown

Homework Equations

The answer is A.

The Attempt at a Solution

Resistance of X = 200^2 / 100 = 400ohm
Resistance of Y= 200^2 / 60 = 667 ohm
Voltage supply to X= 200 (400)/(400+667) =75V
Voltage supply to Y= 200(667)/(400+667)= 125V
From the graph ,put 75 V, I get 0.3 x 75 = 22.5 W for X
put 125 V, I get 125 x 0.2 = 25 W for Y.

Please tell me how to get the correct answers. Thanks for your help.

 

[phyzguy]

Moderator note: Members are reminded that giving complete, nearly complete, or step by step instructions for the solution to a homework question is against forum rules. This post steps over the line but because the OP has already read and quoted it, it has been deemed not worthwhile to remove it now.

The resistances aren't linear, so you can't just treat it as you would linear resistors. You need to use the supplied graph. If the two bulbs are in series, the same current is flowing through both bulbs, and the two voltage drops across the two bulbs need to add up to 200V. So imagine drawing a horizontal line across the graph and sliding it up and down until you find two points that add up to 200V. This gives you the current and the voltage across each bulb, from which you can calculate the power dissipated by each bulb.

 

[Clara Chung]

The resistances aren't linear, so you can't just treat it as you would linear resistors. You need to use the supplied graph. If the two bulbs are in series, the same current is flowing through both bulbs, and the two voltage drops across the two bulbs need to add up to 200V. So imagine drawing a horizontal line across the graph and sliding it up and down until you find two points that add up to 200V. This gives you the current and the voltage across each bulb, from which you can calculate the power dissipated by each bulb.

Thanks. I found the line at 0.25A by your tips.