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Resistor R1 is 12 ohm. Connect it in series with R2 and connect it to 8V power supply. It is known that the voltage at both ends of R2 is 2V. Calculate the current in the circuit
Power supply 8V
So the voltage on R1 V1 = V total - V2 = 8-2 = 6V
So the current is I = V1 / R1 = 6 / 12 = 0.5A
So R2 resistance is R2 = V2 / I = 2 / 0.5 = 4 Ω
Note that because it is in series, the total current in the circuit is the current flowing through the resistors, I = I1 = I2
The resistance values of the two resistors are r1-8ohm and r2-4ohm respectively. They are connected in series to the 24 V power supply. What are the voltages at both ends of R1 and R2
R=R1+R2=8+4=12
I=U/R=24/12=2
U1=R1*I=2*8=16
U2=R2*I=4*2=8
Connect the resistance R1 to the two ends of the power supply whose internal resistance is equal to 1 ohm, and the voltage at both ends of the power supply is measured to be 3V. If a resistance R2 = 6 ohm is connected in series on R1, then connect it to
Connect resistance R1 to both ends of power supply with internal resistance equal to 1 ohm, and the voltage at both ends of power supply is measured to be 3V. If a resistance with R2 = 6 ohm is connected in series on R1, and then connect it to both ends of power supply, and the voltage at both ends is measured to be 4V?
Set the electromotive force of the power supply as E
Then the voltage at the front end of series connection is u = E / (R1 + 1) * R1 = 3
After series connection, terminal voltage U = E / (R1 + 1 + 6) * (R1 + 6) = 4
So R1 = 9 ohm
Given the resistance R1 = 20 ohm and R2 = 30 ohm, if they are connected in series at both ends of the power supply with voltage of u, the ratio of electric heat generated on R1 and R2 is? If they are connected in parallel at both ends of the power supply with voltage of u, the ratio of electric heat generated on R1 and R2 is?
The current of series circuit is equal everywhere, and the heat produced by resistance is equal to
Q = I²R
So R1: R2 produces heat as Q1, Q2
So Q1: Q2 = R1: R2 = 20:30 = 2:3
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