Two bulbs, know the rated voltage and rated power, how to series in the circuit of the actual power How to calculate the actual power in series in the circuit (the circuit voltage is known)?

Two bulbs, know the rated voltage and rated power, how to series in the circuit of the actual power How to calculate the actual power in series in the circuit (the circuit voltage is known)?


P = u * I = u ^ 2 / r = I ^ 2 * r as soon as we see that it is a series circuit, we should think of its basic characteristic that the current is equal everywhere, so we should use the formula about the current: P = I ^ 2 * r now there are two unknowns I and R, and I = u / R total U. we know that the total resistance of the series circuit is the sum of the resistances



Two light bulbs with different power are connected in series on the circuit. Why are the ones with lower power brighter?


There is p = I2R in series, so the resistance of the lamp with low power is small, so when the actual power is large, the lamp will be on



1. In the series circuit, (1) which of the two bulbs with the same rated voltage and different rated power is brighter? (2) the rated voltage is different, and the rated power is the same


In the series circuit, I is the same, P = I2R, so if R is large, the actual power will be large, so the core is to see whose resistance is large
1. If u is the same, it can be concluded from R = u ^ 2 / P that if P is small, R is large, that is, if P is small, R is bright
2. The same as 1, r = u ^ 2 / P, we can see that the R of u is large, that is, the brightness of u is large



Two bulbs with the same rated voltage and different rated power are connected in series. Under their rated voltage, will the bulb with higher rated power be on? Why?


Because P = U2 / R
So r = U2 / P
From the above formula, we can see that the resistance of high power bulb is relatively small
Because it's a series circuit, the circuit is the same
So we use the formula P = I2R
In the case of the same current, the higher the resistance, the greater the power
So we can know that the resistance of the high-power bulb is relatively small
So at this time, its actual power is relatively small, so it is relatively dark