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R1 and rheostat R2 are connected in series, R1 = 10 Ω, the maximum resistance of R2 is 30 Ω, and the power supply voltage is 8 V. When calculating the resistance of rheostat R2 connected into the circuit, R1 consumes What is the maximum electric power of R2? What is the maximum electric power consumed by R1? Can you work out the maximum power of R2
So when R2 = 0, the total resistance is the smallest, I is the largest, and P1 is the largest. At this time, I = u / R1 = 0.8A, so P1 = I & # 178; R1 = 6.4wp2 = I & # 178; R2 = [U / (R1 + R2)] &# 178; R2 = u & # 178; R2 / (R1 + R2) &# 178; = u & # 178; / [(r2-r1) / r2 + 4r1] because u and R1 are fixed, so
Resistor R1 = 4 ohm, R2 = 8 ohm, connected in parallel to 4 V circuit, the total current is_____ For a total current of 3 A, replace R2 with one______ The resistance of ohm and R1 are connected into the circuit
1 / R total = 1 / R1 + 1 / r2
So r is always 8 / 3
So I = u / r = 3 / 2A
Second, it's easy to ask,
R = u / I = 4 / 3 Ω
1 / R total = 1 / R1 + 1 / r2
So R2 = 12 / 5
Oral calculation, you can calculate by yourself, anyway, the formula is right, physics is my strong point
Two resistors of resistor R1 and R2 are connected in series in the circuit. When the total voltage at both ends of them is u, the current is expressed as I
In this paper, another resistor R is used instead of R1, R2 is connected in the circuit, and the voltage U remains the same. Try to deduce: if the current representation number remains the same, the condition r should satisfy is r = R1 + R2
(necessary formula, including necessary explanation.)
To deduce this formula, we need to take the relationship between neutral and voltage in series circuit
Because the series circuit has partial voltage, that is, u = U1 + U2
And u = IR, substituting into the above formula, IR = i1r1 + i2r2
Because the current in series circuit is equal everywhere, that is, I = I1 = I2
R=R1+R2
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