R1 = 20 ohm, the total resistance of the circuit is 12 ohm, and the current indication is 0.3A?

R1 = 20 ohm, the total resistance of the circuit is 12 ohm, and the current indication is 0.3A?

1 / R1 + 1 / r2 = 1 / R total
1 / 20 + 1 / r2 = 1 / 12, R2 = 30 Ω

It is known that R1 = 6, the maximum allowable current is 2a, R2 = 2 Ω, and the maximum allowable current is 3a
1. Connect r1r2 in parallel, the total current through r1r2 is 2a, and calculate the power supply voltage
2. When r1r2 is connected in parallel, what is the maximum current of trunk circuit?
3. If r1r2 is connected in series, what is the maximum total voltage at both ends of the circuit?

(1) R = r1r2 / (R1 + R2) = 1.5 Ω u = RI = 3V
（2）I=I1+I2=2+3=5A
(3) Because R1 = 6, the maximum current allowed to pass is 2a, resistance R2 = 2 Ω, the maximum current allowed to pass is 3a
R = R1 + R2 = 8 Ω, so u = 16V

R1 = 10 ohm, R2 = 20 ohm. After closing the switch, the current indication is 0.3A. What is the voltage at the two ends of resistor R1? What is the current through R2?
I'm sorry I don't have a picture

They are 3V, 0.3A in series and 2V, 0.1A in parallel

In 2 minutes, 30 banks of electricity pass through a resistor with a resistance value of 100 ohm. The resistor does electric work of joules, and the resistor consumes electric energy of joules

First, the current I = q / T = 30C / 120s = 0.25A in the resistance is calculated
R = 100 Ω,
W = I ^ 2rt = 0.25A × 0.25A × 100 Ω × 120s = 750j
In addition, the problem should be for the current to do 750 J of work, do how much work to consume how much energy, so consumed 750 joules of electricity
DUI AI=q/t=30/2*60=0.25A r=100 W=I^2*rt=0.25*0.25*100*120=750J

The resistance of an electric bell is 12 Ω and the rated voltage is 6V. Now there is only one 18V power supply. In order to make the electric bell work normally, it is necessary to connect a certain resistance. (1) draw the circuit diagram. (2) calculate the resistance of the circuit to be connected

(1) The power supply voltage is greater than the rated voltage of the electric bell. In order to make the electric bell work normally, a voltage divider should be connected in series to the electric bell. The circuit diagram is shown in the figure. (2) when the electric bell works normally, u electric bell = 6V, ∵ the current in the series circuit is equal, ∵ according to Ohm's law, the current in the circuit: I = I electric bell = u electric bell, R electric bell = 6v12, Ω = 0.5A, ∵ in series The total voltage in the circuit is equal to the sum of the partial voltages, the voltage at both ends of the resistor: ur = U-U, bell = 18v-6v = 12V, the resistance value of the series resistor: r = URI = 12v0.5a = 24 Ω. Answer: (1) the circuit diagram is shown in the figure; (2) a 24 Ω resistor should be connected in series

An incandescent lamp marked with "12V, 36W" should be connected to the circuit with 18V power supply voltage. In order to make the bulb shine normally, what resistance should be connected in series? (draw a sketch according to the title)

A:
First of all, to make the bulb light normally, the current is consistent with the voltage is consistent with the line, the following are calculated
From P = UI, I = P / u = 36W / 12V = 3A, so the current in the series circuit is 3a, and the resistance is 3a
The voltage on the lamp is 12V, that on the resistor is (18v-12v = 6V)
So the resistance R = u / I = 6V / 3A = 2 Ω
I hope I can help you. Thank you

There is a small light bulb marked "8V & nbsp; & nbsp; 2W". If it is connected to 12V power supply, it must be connected to the power supply___ Connect a resistance to___ The resistance of the small bulb can make it shine normally

When the bulb works normally, the current is I = Pu = 2w8v = 0.25A. If it is connected to 12V power supply, the voltage at both ends of the bulb should be 8V, then a resistor in series is needed in the circuit. The voltage at both ends of the resistor is ur = u-ul = 12v-8v = 4V, and the resistance value in series is r = URI = 4v0.25a = 16 Ω

To connect a small bulb marked with "12V 0.5A" to the 18V power supply, how large a resistor must be connected in series to make the bulb shine normally?
What is the power of the resistor? What are the advantages and disadvantages of doing so?

I=0.5A
R = (U-U) / I = (18-12) / 0.5 = 12 Ω
Connect a 12 Ω resistor in series to make the bulb shine normally
P = I (U-U) = 0.5 * (18-12) = 3W
The power of this resistor is 3W
Advantages: the bulb voltage and current can meet the rated voltage and current
Shortage: Waste 3W of electric power

A 12V, 6W bulb, if you want to connect it to 18V power supply, you should () fill in series or parallel a () ohm resistor

R light = u ^ 2 / P = 12 ^ 2 / 6 = 24
I = P / u = 6 / 12 = 0.5
R total = u '/ I = 18 / 0.5 = 36
R'=36-24=12
Series 12 Euro process unit, I wrote it

The resistance of an electrical appliance is 120 ω. To make a quarter of the total current in the circuit pass through this appliance, thank you,
The resistance of an electrical appliance is 120 ω. To make a quarter of the total current in the circuit pass through the appliance, connect a resistance of () ω with the appliance (). If the voltage at both ends of the appliance is one fourth of the total voltage, connect a resistance of () ω with the appliance ()

The resistance of an electrical appliance is 120 ω. To make a quarter of the total current in the circuit pass through the appliance and connect a (40) ω resistor with the appliance. If the voltage at both ends of the appliance is a quarter of the total voltage, a (360) ω resistor should be connected with the appliance (in series)