Question

Задача. На цоколі однієї лампочки написано «12 В; 6 Вт», на цоколі іншої «12 В; 4 Вт». Лампочки зʼєднані послідовно. Яку максимальну напругу можна подати на лампочки, щоб вони не перегоріли? Яка потужність при цьому виділиться в лампочках? ПОМОГИТЕ ПОЖАЛУЙСТА! ДАЮ 100 БАЛЛОВ!

Answer 1

Ответ: максимальный ток 1/3 А. Остальное по закону Ома.

Answer 2

Problem Analysis

The problem states that there are two light bulbs connected in series. The first bulb has a voltage rating of 12V and a power rating of 6W, while the second bulb has a voltage rating of 12V and a power rating of 4W. We need to determine the maximum voltage that can be applied to the bulbs without causing them to burn out, as well as the power dissipated in the bulbs under this condition.

Solution

When light bulbs are connected in series, the total voltage across the bulbs is divided between them according to their resistance or power ratings. In this case, since the bulbs have the same voltage rating, the voltage across each bulb will be the same.

To find the maximum voltage that can be applied to the bulbs without causing them to burn out, we need to consider the power ratings of the bulbs. The power dissipated in a bulb can be calculated using the formula:

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

Since the bulbs are connected in series, the current passing through both bulbs will be the same. Therefore, we can rewrite the formula as:

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

Since the power ratings of the bulbs are given, we can calculate the current passing through the bulbs using the formula:

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

Let's calculate the current passing through the bulbs using the power ratings provided:

For the first bulb: Power (P1) = 6W Voltage (V1) = 12V Current (I1) = P1 / V1

For the second bulb: Power (P2) = 4W Voltage (V2) = 12V Current (I2) = P2 / V2

Since the bulbs are connected in series, the total current passing through both bulbs will be the sum of the currents passing through each bulb:

Total Current (Itotal) = I1 + I2

Now, let's calculate the total current passing through the bulbs:

Itotal = I1 + I2

Finally, the maximum voltage that can be applied to the bulbs without causing them to burn out is the sum of the voltage ratings of the bulbs:

Maximum Voltage = V1 + V2

Let's calculate the maximum voltage and the power dissipated in the bulbs:

Calculation

Given: Voltage rating of the first bulb (V1) = 12V Power rating of the first bulb (P1) = 6W Voltage rating of the second bulb (V2) = 12V Power rating of the second bulb (P2) = 4W

Calculations: Current passing through the first bulb (I1) = P1 / V1 Current passing through the second bulb (I2) = P2 / V2 Total current passing through the bulbs (Itotal) = I1 + I2 Maximum voltage that can be applied to the bulbs = V1 + V2 Power dissipated in the bulbs = Itotal * Maximum voltage

Substituting the given values into the formulas:

I1 = 6W / 12V I2 = 4W / 12V Itotal = I1 + I2 Maximum Voltage = 12V + 12V Power dissipated in the bulbs = Itotal * Maximum Voltage

Now, let's calculate the values:

I1 = 0.5A I2 = 0.333A Itotal = 0.5A + 0.333A Maximum Voltage = 12V + 12V Power dissipated in the bulbs = Itotal * Maximum Voltage

I1 = 0.5A I2 = 0.333A Itotal = 0.833A Maximum Voltage = 24V Power dissipated in the bulbs = 0.833A * 24V

Therefore, the maximum voltage that can be applied to the bulbs without causing them to burn out is 24V, and the power dissipated in the bulbs under this condition is 19.992W.

Conclusion

To summarize, when two light bulbs with voltage ratings of 12V and power ratings of 6W and 4W, respectively, are connected in series, the maximum voltage that can be applied to the bulbs without causing them to burn out is 24V. Under this condition, the power dissipated in the bulbs is approximately 19.992W.