When the voltage at both ends of a fixed value resistor changes from 6V to 10V, the current flowing through the resistor increases by 0.1A______ Ω．

When the voltage at both ends of a fixed value resistor changes from 6V to 10V, the current flowing through the resistor increases by 0.1A______ Ω．

Let R be the resistance of the constant value resistor. From the meaning of the question, we can get 6vr + 0.1A = 10vr, and the solution is r = 40 Ω

As shown in the figure, the electromotive force of the power supply e = 12V, the internal resistance R = 1 Ω, the resistance R1 = 3 Ω, R2 = 2 Ω, R3 = 5 Ω, the capacitance of the capacitor C1 = 4 μ F, C2 = 1 μ F, calculate the charged quantity of C1 and C2

When the circuit is stable, resistor R1 and R2 are connected in series, the voltage of capacitor C1 is & nbsp; U1 = r2r2 + R1 + re = 22 + 3 + 1 × 12V = 4V, the voltage of capacitor C2 is & nbsp; U2 = R1 + r2r1 + R2 + re = 10V, the charge of capacitor C1 is Q1 = c1u1 = 4 × 10-6 × 4C = 1.6 × 10-5c & nbsp; Q2 = c2u2 = Q2 = 1 × 10-6 × 10C = 1 × 10-5c answer: the charge of C1 and C2 is 1.6 × 10-5c and 1 × 10-5c respectively

As shown in the figure, e = 10V, the internal resistance of the power supply is ignored, R1 = 4 Ω, R2 = 6 Ω, C1 = C2 = 30 μ F. first close the switch s, and then open s after the circuit is stable, and calculate the amount of electricity flowing through the resistance R1 after opening s

When the switch is closed, the current in the circuit is I = ER1 + R2 = 10v4 Ω + 6 Ω = 1A; the voltage at both ends of R2 is u = IR2 = 6V; therefore, the quantity of charge on capacitor C1 is Q1 = UC = 6V × 30 μ f = 1.8 × 10-4c; when the switch is disconnected, the two capacitors are directly connected with the power supply, and the voltage is equal to the electromotive force of the power supply; therefore, the total quantity of charge on the two capacitors is q =

As shown in the figure, if R1 = 3 Ω and R2 = 6 Ω, the resistance between a and B is______ And______ ．

When the switch is off, the two resistors are in parallel, so 1R = 1r1 + 1r2, that is, r = 2 Ω; when the switch is closed, the two resistors are short circuited, so the resistance between AB is 0 Ω; so the answer is: 2 Ω, 0 Ω