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The resistance R1 and R2 are connected in series in the circuit, the voltage at both ends of R1 is 3V, R2 voltage is 9V, R1 + R2 = 600 Ω
① In a series circuit, the total voltage is equal to the sum of the voltages at both ends of each consumer. U = U1 + U2 = 3V + 9V = 12V,
② According to r = u ﹣ I, I = u ﹣ r = 12V ﹣ 600 = 0.02A
③ According to r = u △ I, R1 = U1 △ I = 3V △ 0.02A = 150, R2 = U2 △ I = 9V △ 0.02A = 450
Why does the external circuit consume the most power when the external resistance is equal to the internal resistance?
Use Ohm's law of closed circuit
Output power of power supply: refers to the electric power of external circuit, which is equal to the product of total current I and terminal voltage u in quantity
P =IU
That is, for the pure resistance circuit, the output power of the power supply
P=I2R=(E/R+r)2R=E2R/((R-r)2+4Rr)=E2/((R-r)2/r+4r)
It can be seen from the above formula that when the external resistance is equal to the internal resistance of the power supply (r = R), the output power of the power supply is the largest, and its maximum output power is E2 / 4R
4. When the voltage applied to both ends of the fixed value resistor increases from 8V to 12V, the current through the resistor changes by 0.1A correspondingly, and the electric power consumed by the fixed value resistor changes by 0.1A
A、0.4W B、0.8W C、2.0W D、3.6W
C
12/R-8/R=0.1
R = 40 Ω
12 times 12 divided by 40 - 8 times 8 divided by 40 = 2.0
The voltage on both sides of resistance R is increased from 2V to 6V, and the current is increased by 0.1A. How much is the electric power increased?
2/I=6/(I+0.1 I)
I=0.05A
R = 2V / 0.05A = 40 Ω
P2-P1=6V*0.15A-2V*0.05A=0.8w
That is, the electric power increases by 0.8W
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