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As shown in the figure, U1 = 6V, u = 10V, the current passing through R1 is 0.4A, and the resistance of R1 R2 can be calculated The circuit diagram is patchy___________ I ————|_________ |——————|___________ |————＞—— R1 R2 |＜——U1——＞| |————U2——————＞| |＜——————————U————————————＞|
R1 current = R2 current = 0.4A, each current in series circuit is equal
U2=U-U1=10-6=4V
R1 = U1 / I1 = 6V / 0.4A = 15 Ω
R2 = U2 / I2 = 4V / 0.4A = 10 Ω
In the figure, what is the relationship between the voltage ratio U1 / u and the resistance ratio R1 / R1 + R2?
Do not see figure, but it can be seen that this is a series circuit, otherwise there is no voltage ratio
If U1 is the voltage at both ends of R1 and u is the total voltage, there are only two resistors r1r2 connected in series in the circuit
It can be seen that U1 / u is the ratio of R1 partial voltage to total voltage, and R1 / R1 + R2 is the ratio of R1 resistance to total resistance
According to the principle of series connection, U1 / u = R1 / R1 + R2
In the circuit shown in the figure, the power supply voltage is 6V, the resistance of R2 is 10 Ω, and the current through R1 is 0.4A
Find: [1] resistance of R1; [2] current through resistance R2; [3] current I in trunk circuit
This is a problem in the exercise book of grade two in junior high school. In addition, I won't take a screenshot
It can be seen that the figure should be two resistors in parallel
R1=U/I1=6v/0.4A=12.5Ω
I2=U/R2=6v/10Ω=0.6A
I total = I1 + I2 = 0.4 + 0.6 = 1A
RELATED INFORMATIONS
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