The bulb "pz220-15" and the bulb "pz220-40" are connected in series to calculate the maximum current and maximum voltage in the circuit?

The bulb "pz220-15" and the bulb "pz220-40" are connected in series to calculate the maximum current and maximum voltage in the circuit?

According to P = a * V V = R * a, the current of "pz220-15" is 0.068, the resistance is 3235 Ω, the current of "pz220-40" is 0.18, and the resistance is 1222 Ω
The maximum current is "pz220-15" and the current is 0.068a
The maximum voltage is 220 + 0.068 * 1222 = 303v

In a residential area near a substation, the user voltage is often 22V higher than the normal voltage. If the pz220-40 bulb is connected to the circuit, the maximum actual power of the bulb is

According to P = u & # 178 / R, we know that P is proportional to u & # 178; when R is constant
P1/P2=U1²/U2²
40w/P2=(220V)²/(220V+22V)²
P2=48.4w
That is, the maximum actual power of the bulb is 48.4w

A bulb marked with "pz220-40" is connected to a 110V circuit. What is the actual current through the lamp?

You can use "pz220-40" to calculate the resistance, r = u ^ 2 / P, and then use Ohm's law to calculate the current: I = u / R;
U ^ 2 is the square of U. r = u ^ 2 / P is derived from P = UI and Ohm's law. It's very important. I will tell you

A bulb marked with pz220-25, (1) rated current under normal operation, (2) how much electric energy does it consume when it works for 20 minutes
3) How many kilowatt hours of electricity are consumed every month when using 4 hours a day? (in 30 days)（

（1）
P=UI
I=P/U=25/220=0.114A
（2）P=w/t
W=Pt,P=25W,t=20min=1200s
=25×1200
=30000J
（3）W=Pt,P=25w=0.025kw ,t=4×30＝12h
=0.025×120
=3kw·h
That is 3 degrees of electricity
From the "middle school physics navigation" answer