A capacitor and a resistor are connected in series on the 220 V AC power supply. It is known that the voltage drop on the resistor is 120 V, so what is the voltage on the capacitor

Uc=√(220²-120²)=184.39V

There is a small bulb, which is connected in parallel with a 10 ohm resistor in a constant voltage circuit. When the main current is 0.6A, the small bulb happens to work normally
At this time, its power is 0.8W

Let the power supply voltage be u and the bulb resistance be r, then r = (U squared) / P = (U squared) / 0.8W. Substituting U / R plus U / 10 = 0.6A, we can get (0.8W / U) plus (U / 10) = 0.6A, then U1 = 2V, U2 = 4V, because r = (U squared) / P, when U1 = 2V, R1 = 5ohm, when U2 = 4V, R2 = 20ohm

In the circuit shown in the figure, the resistance value of resistance R1 is 5 Ω, the current I in the trunk circuit is 0.5A, and the voltage at both ends of resistance R2 is 2V

(1) Because two resistors are in parallel: & nbsp; & nbsp; U1 = U2 = 2V, according to Ohm's law, I1 = u1r1 = 2v5, Ω = 0.4A; & nbsp; & nbsp; & nbsp; & nbsp; (2) according to the current law of parallel circuit, I2 = i-i1 = 0.5a-0.4a = 0.1A; & nbsp; & nbsp; (3) the voltage at both ends of R2 is 2V

R1 = 20 Ω, R2 = 10 Ω. If two resistors are connected in series, what is the maximum voltage at both ends of the circuit? If two resistors are connected in parallel, what is the maximum current in the trunk circuit
The maximum allowable current of R2 is 2A. The maximum allowable voltage of R1 is 30V

In series: because the maximum allowable voltage at both ends of R1 is 30V, the maximum allowable current is 30 / 20 = 1.5A. When two resistors are in series, the maximum allowable voltage at both ends of the circuit is (20 + 10) * 1.5 = 45V

A capacitor and a resistor are connected in series on the 220 V AC power supply. It is known that the voltage drop on the resistor is 120 V, so what is the voltage on the capacitor