6. In the circuit shown in Figure 2, the internal resistance of the power supply can not be ignored. It is known that the setting resistance R1 = 10 Ω, R2 = 8 Ω. When the single pole double throw switch S is placed in position 1

6. In the circuit shown in Figure 2, the internal resistance of the power supply can not be ignored. It is known that the setting resistance R1 = 10 Ω, R2 = 8 Ω. When the single pole double throw switch S is placed in position 1

.1.9V
According to the principle of series voltage division, terminal voltage U1 = 2 = 10e / (10 + R) U2 = 8e / (8 + R)
The results show that U2 = 1.6 (10 + R) / (8 + R)
That is, U2 = 1.6 (1 + 2 / (8 + R))
That is, U2 = 1.6 + 3.2 / (8 + R)
U2 = 2V at r = 0 and U2 = 1.6V at R infinity
It can only be between the two

In the circuit shown in the figure, the power supply voltage is 6V and remains unchanged, and the resistance values of R1, R2 and R3 are 8 Ω, 4 Ω and 12 Ω respectively? (2) If the switches S1 and S2 are closed, what is the indication of the ammeter?

(1) If the switches S1 and S2 are disconnected, R1 and R2 are connected in series, the voltmeter measures the voltage at both ends of R2, and the ammeter measures the current in the circuit. Because the total resistance in the series circuit is equal to the sum of the partial resistances, the indication of the ammeter in the circuit is: I = ur1 + R2 = 6v8 Ω + 4 Ω = 0.5A, the indication of the voltmeter is: U2 = IR2 = 0.5A × 4 Ω = 2V; (2) if the switches S1 and S2 are closed, R1 and R3 are connected in parallel, the ammeter is closed To measure the main circuit current, because the reciprocal of the total resistance in the parallel circuit is equal to the sum of the reciprocal of each resistance, the total resistance in the circuit: r = r1r3r1 + R3 = 8 Ω × 12 Ω, 8 Ω + 12 Ω = 4.8 Ω, and the indication of the main circuit ammeter: I '= ur = 6v4.8 Ω = 1.25A The indication of the flow table is 1.25A

R1 = 10 Ω, the maximum resistance value of R2 is 30 Ω, and the power supply voltage value is 8 V. When calculating the resistance of the rheostat R2 connected to the circuit, what is the maximum power consumption on R2?

Because of series connection, the current is equal everywhere I = u / (R1 + R2)
Using P = I2R
Substituting the expression of current into p-transformer = I2R (the resistance inside is R2),
We can get p-variable (maximum) = U2 / 4r1 = 1.6W (the maximum power can be obtained by taking the numerator and denominator at the same time as R2 and using ab with a + b greater than or equal to 2 times the root sign)
Attachment: the deformation process is too complicated here. It seems that there is something wrong with sending pictures in the past two days. Sorry, the premise for the maximum power of rheostat is R1 = R2

The power supply voltage of the series circuit is 6V, the resistance R1 = 2 Ω, and the voltage at both ends of R2 is 4V. Calculate the resistance of R2

U = 6 V, R1 = 2 Ω, U2 = 4 V
Analysis: from the meaning of the title, the series circuit is two resistors R1 and R2 in series
So the voltage at both ends of R1 is U1 = u-u2
And U1 / R1 = U2 / R2
That is, (u-u2) / R1 = U2 / R2
（6－4）/ 2＝4 / R2
R2 = 4

In the circuit shown in the figure, the current representation is 0.3A, the resistance R1 = 40 Ω, R2 = 60 Ω, and the main circuit current I

Resistor R1 and R2 are in parallel, ∵ I1 = 0.3A, R1 = 40 Ω, ∵ U1 = i1r1 = 0.3A × 40 Ω = 12V, it can be seen that the voltage at both ends of R2 is: U2 = 12V, R2 = 60 Ω, ∵ the current through resistor R2 is: I2 = u2r2 = 12v60 Ω = 0.2A, then the total current in the trunk is: I = I1 + I2 = 0.3A + 0.2A = 0.5A