Resistance R1 = 8 ohm, resistance R2 = 10 ohm, power supply voltage and setting resistance R are unknown, When the switch is connected to position 1, the indication of ammeter is 0.2A. When the switch is connected to position 2, what is the possible value of indication? When the switch is connected to position 1, there are only three of them in the circuit, resistance R1, constant resistance R, ammeter, which are in series; when the switch is connected to position 2, there are only three of them in the circuit, which are in series!

Resistance R1 = 8 ohm, resistance R2 = 10 ohm, power supply voltage and setting resistance R are unknown, When the switch is connected to position 1, the indication of ammeter is 0.2A. When the switch is connected to position 2, what is the possible value of indication? When the switch is connected to position 1, there are only three of them in the circuit, resistance R1, constant resistance R, ammeter, which are in series; when the switch is connected to position 2, there are only three of them in the circuit, which are in series!

1. When s is disconnected, there is no current in the circuit, and the charged quantity Q = EC
When s is closed, the current I = E / (R + R1 + R2) = 1 / 6
Voltage at both ends of capacitor u = IR2 = 5
So the ratio of electric quantity = E / u = 2
2. When closed, the charge passing through R3 = UC / 2 = 10-7 Coulomb

It is known that the resistance R1 is equal to 10 ohm and the resistance R2 is equal to 20 ohm. Connect R1 and R2 in series and then connect them to a 6V power supply
Voltage at both ends of R1

R1 = 10 Ω, R2 = 20 Ω, so the total resistance in series circuit R = 30 Ω;
Total voltage U = 6V,
According to Ohm's law, u = IR,
So the voltage at both ends of R1 U1 = (R1 / R) u = (10 / 30) x 6V = 2V

After resistor R1 and R2 are connected in series, they are connected to the power supply with constant voltage for 1 minute. It is known that R1 is 5 ohm, R2 is 10 ohm, and the voltage at both ends of R1 is 2V. Calculate the relevant physical quantities

I=U/(R1+R2)=2/15A
U1=IRI=2/15*5=2/3V
U2=IR2=2/15*10=4/3V
Calculate the power and electric energy by yourself, and turn minutes into hours
P=I^2R
W=UIT

When a power supply is connected with 8 ohm resistance, the current passing through the power supply is 0.15A, and when it is connected with 13 ohm resistance, the current passing through the power supply is 0.10a. Calculate the electromotive force and internal resistance of the power supply

I haven't done this for a long time! Try it!
R0 = internal resistance
U = electromotive force
U U
--------=0.15 ------- =0.10
R0+8 R0+13
0.15R0+1.2=0.1R0+1.3
The results are as follows
R0 = 2 Ω
U=1.5V
I don't know whether it's true or not!