A constant resistance is marked with 10 Ω 1 A. if it is connected to both ends of a 12 V circuit and works normally, how to connect a large resistance in the circuit? A constant resistance is marked with "10 Ω 1 a". If it is connected to both ends of a 12 V circuit and works normally, how to connect a large resistance in the circuit? If it is used in parallel with a resistor marked "20 Ω 1.5 a", what is the maximum voltage that can be applied to the circuit?

A constant resistance is marked with 10 Ω 1 A. if it is connected to both ends of a 12 V circuit and works normally, how to connect a large resistance in the circuit? A constant resistance is marked with "10 Ω 1 a". If it is connected to both ends of a 12 V circuit and works normally, how to connect a large resistance in the circuit? If it is used in parallel with a resistor marked "20 Ω 1.5 a", what is the maximum voltage that can be applied to the circuit?

one
The voltage at both ends of series voltage divider R is u = 12-10 = 2V
Over R current I = 1A
R = u / I = 2V / 1A = 2 Ω
two
U = 1 * 10 = 10V
U parallel = 20 * 1.5 = 30V
The applied voltage U is the smaller of the rated voltages of the two resistors
U=10V

The resistance of an electrical appliance is 60 ohm. If you want to use 1 / 5 of the total current in the circuit to pass through the electrical appliance, what resistance should be connected in series

It's better not to connect in series, which will lead to partial voltage (only 1 / 4 of the original) and your electrical appliances can't work normally
Better get a shunt

4. The rated power of an electric appliance with 24 ohm resistance is 6 watts. In order to make it work normally when it is connected to a 20 volt power supply, what is the resistance in series in the circuit
4. The rated power of a consumer with 24 ohm resistance is 6 watts. In order to make it work normally when it is connected to a 20 volt power supply, what is the resistance in series in the circuit? How much power does this resistor consume in one hour?

From P = UI and I = u / R, u = √ (PR) is obtained to calculate the rated voltage of electrical appliances
Uo=√(PoRo)=√(6WX24Ω)=12V；
I = u / R is used to calculate the normal working current of electric appliance
Io=Uo/Ro=12V/24Ω=0.5A；
R = u / I is obtained from I = u / R, and the resistance to be connected in series is calculated
R'=U'/Io=(U-Uo)/Io=(20V-12V)/0.5A=16Ω；
From w = uit and I = u / R, we can get w = u ^ 2T / R to calculate the power consumption of series resistance
W'=U'^2t/R'=(8V)^2X3600s/16Ω=14400J

If the current in the circuit is 0.5 A, calculate the total resistance? Supply voltage? Current ratio? Two resistance voltage ratio

Because it is in series, the total resistance is 9 Ω, and the supply voltage U = 0.5 times 9 = 4.5 v
Because the series current is equal everywhere, the ratio of current is 1:1. Because it is in series, the ratio of voltage to resistance is 3 ohm to 6 ohm = 1:2