In the circuit diagram shown in the figure, R1 = 9 Ω, R2 = 15 Ω, R3 = 30 Ω, The electromotive force E of the battery pack is 12 V, and the internal resistance R is 1 ohm. Calculate the reading of the ammeter. "The picture shows that R1 is connected in series with the ammeter, then connected in parallel with R3, and finally connected in series with R1. (R1 is on the trunk road)"

In the circuit diagram shown in the figure, R1 = 9 Ω, R2 = 15 Ω, R3 = 30 Ω, The electromotive force E of the battery pack is 12 V, and the internal resistance R is 1 ohm. Calculate the reading of the ammeter. "The picture shows that R1 is connected in series with the ammeter, then connected in parallel with R3, and finally connected in series with R1. (R1 is on the trunk road)"

According to the meaning of the title, the picture is as follows:
@ ___ R1__________
@ | @@@@@@@@@@ |@|
@ | @@@@@@@@@@ R3 R2
@ | @@@@@@@@@@ |@|
@ r @@@@@@@@@@ |@A
@ | @@@@@@@@@@ |@|
@_ |__ @@@@@@@@ |@|
@ __ @@@@@@@@@ |@|
@ |____________ |_|
Total resistance R = R + R1 + R2 * R3 / (R2 + R3)
Total current of trunk road I = E / R
R2, R3 shunt, R2 branch current value is
I2 = I*R3/(R2+R3)
The reading of ammeter is I2, i.e
E/(r + R1 + R2*R3/(R2+R3)) * R3/(R2+R3)
=E*R3/(R2R3+(R1+r)(R2+R3))
=12*30/(15*30+10*45)=0.4 A

As shown in the figure, connect R1 and R2 in series in the circuit. The resistance value of R1 is 12 Ω. The power supply voltage U measured by Voltmeter is 8 V, and the voltage U2 at both ends of R2 is 2 v. please use two different methods to calculate the resistance value of R2,

R1 and R2 in series
R1 terminal voltage U1 = u-u2 = 8v-2v = 6V
In the circuit, the current I = U1 / R1 = 6V / 12ohm = 0.5A
The resistance of R2 is R2 = U2 / I = 2V / 0.5A = 4 Ω
Method 2
R1 and R2 in series
R1 terminal voltage U1 = u-u2 = 8v-2v = 6V
In the circuit, the current I = U1 / R1 = 6V / 12ohm = 0.5A
Total resistance R = u / I = 8V / 0.5A = 16 Ω
The resistance of R2 is R2 = r-r1 = 16 Ω - 8 Ω = 4 Ω

In the parallel circuit (only two resistors in parallel), when the resistance values of the two resistors change, but R1 + R2 = R is the fixed value (for example, R1 is increased by 10 Ω,

Again:
1 / R equivalent = 1 / R1 + 1 / r2 = 1 / R1 + 1 / (r-r1) = R / R1 (r-r1)
R equivalent = R1 (r-r1) / R
From the basic inequality ab

With a 40 ohm resistor, the maximum current allowed to pass is 0.2A. If it is connected to a 20V power supply, the resistor can work normally? What is the resistance in series?

20 / (R + 40) = 0.2, r = 60 Ω