I can help you with your question about power dissipation in parallel circuits.
To find the ratio of the resistances of the two light bulbs, we can use the formula for power dissipation in a parallel circuit:
P = V 2 R {\displaystyle P={\frac {V^{2}}{R}}}
where P is the power, V is the voltage, and R is the resistance.
Since the two light bulbs are connected in parallel, they have the same voltage across them. Let’s call this voltage V. Then we can write:
P 1 = V 2 R 1 {\displaystyle P_{1}={\frac {V^{2}}{R_{1}}}}
P 2 = V 2 R 2 {\displaystyle P_{2}={\frac {V^{2}}{R_{2}}}}
Dividing these two equations, we get:
P 1 P 2 = R 2 R 1 {\displaystyle {\frac {P_{1}}{P_{2}}}={\frac {R_{2}}{R_{1}}}}
Now we can plug in the given values of P1 and P2 and solve for the ratio of R2 to R1:
30 40 = R 2 R 1 {\displaystyle {\frac {30}{40}}={\frac {R_{2}}{R_{1}}}}
0.75 = R 2 R 1 {\displaystyle 0.75={\frac {R_{2}}{R_{1}}}}
Therefore, the ratio of their resistances is 0.75. This means that the light bulb with higher power dissipation has lower resistance, and vice versa.