The power supply voltage is 6V, the resistance R1 = 5 ohm, and the maximum resistance of rheostat R2 is 10 ohm Find out the change range of voltmeter and ammeter

The power supply voltage is 6V, the resistance R1 = 5 ohm, and the maximum resistance of rheostat R2 is 10 ohm Find out the change range of voltmeter and ammeter

When the resistance of R2 is 0, the reading of ammeter is 6 △ 5 = 1.2A, and the reading of voltmeter is 1.2 × 5 = 6V
When the resistance of R2 is 10, the reading of ammeter is 6 △ 15 = 0.4A, and the reading of voltmeter is 0.4 × 5 = 2V
The range of ammeter is 0.4A ~ 1.2A, and the range of voltmeter is 2V ~ 6V

Connect R1 = 2 ohm resistance and R2 = 4 ohm resistance in series to the power supply
Connect R1 = 2 ohm resistor and R2 = 4 ohm resistor in series to the power supply. The voltage at both ends of R1 is known to be 1V. Find (1) power supply voltage; (2) what are the electric power of R1 and R2?

1) 3V, series resistance ratio equals voltage ratio, 1 + 2 = 3
2）p=u^2/r,Pr1=0.5w,Pr2=1w

Resistance R1 = 10 ohm, R2 = 20 ohm. If it is connected in series in a circuit with voltage of 6V, what is the voltage at both ends of R1
Why?

Because the total resistance in the series circuit R = R1 + R2
So the total resistance R = 10 + 20 = 30R
And because the total voltage is 6V
So I = u / r = 6V / 30R = 0.2A
So the voltage of R1 is IR1 = 0.2A * 10 ohm = 2V

If U1: U2 = 2:3 And R1 = 20 ohm, then R2 is equal to (30) ohm,
If r1r2 is connected to both ends of the original circuit in parallel, why is the ratio of the total current in series to that in parallel (6:25)?

Series current I = u / (R1 + R2) = u / 50
Parallel total current I parallel = u / R1 + U / r2 = (R1 + R2) U / r1r2 = 50U / 600
I string: I parallel = 6:25