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If two light bulbs were connected in parallel with each other's and with a certain battery. The powers dissipated by them are P1 = 30 Watts and P2 = 40 Watts, respectively. The ratio of their resistances (R1/R2) is: 1.

User Jashira
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2 Answers

4 votes

Answer:

R1 / R2 = 4/3

Step-by-step explanation:

In a parallel circuit, the voltage across each component is the same, but the current through each component can be different. The power dissipated by a component in an electrical circuit can be calculated using the formula:

Power (P) = Voltage (V) * Current (I)

Since the voltage across both light bulbs is the same, we can set up the following equation:

P1 = V * I1
P2 = V * I2

Divide the two equations to find the ratio of their resistances:

P1 / P2 = (V * I1) / (V * I2)

Since V is common in both terms, it cancels out:

P1 / P2 = I1 / I2

Now, we know the power dissipated by each bulb:

P1 = 30 Watts
P2 = 40 Watts

Substitute these values into the equation:

30 / 40 = I1 / I2

Simplify the ratio:

3/4 = I1 / I2

Since the current is inversely proportional to resistance in a parallel circuit, the ratio of their resistances (R1 / R2) will be the reciprocal of the ratio of their currents:

R1 / R2 = 1 / (I1 / I2) = 1 / (3/4) = 4/3

So, the ratio of their resistances (R1 / R2) is 4/3.
User Fjarlaegur
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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.

User Mrbellek
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