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DC power supply 5V need resistance to step down to 4V, limit current in 4.5a need a few w several r resistance?
The partial voltage of series resistance is: U2 = u-u1 = 5-4 = 1 (V)
Because the current limit is 4.5a, i.e. I2 = 4.5a
Then R2 = U2 / I2 = 1 / 4.5 ≈ 0.22 (Ω)
Resistance power: P = i2u2 = 4.5 × 1 = 4.5 (W)
A: slightly
Hope to help you, if you have any questions, you can ask~~~
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Voltage 5V, need to be reduced to 4V. How much resistance do you need
Using resistor to reduce voltage needs to know the working current. It is recommended that you connect rectifier diodes in series, which is relatively simple and reliable. The forward voltage drop of a diode is 0.0.7v, and 1 ~ 2 diodes in series can be determined experimentally
When 3V voltage is applied at both ends of a certain resistance, the current passing through the conductor is 0.2A. When the current passing through the conductor increases by 0.2A, what is its resistance
The original resistance is 15 ohm; when the current through the conductor increases by 0.2A, its resistance is: 15 / 2 = 7.5 ohm. According to the Ohm's law of DC circuit: I = V / R, it is very easy to find
Two bulbs are connected in parallel, the power supply voltage is 3V, the current of bulb is 0.6A, the resistance of L2 is 10 Ω
Known: u = 3V; I1 = 0.6A; R2 = 10 Ω ask: I =? Because it is a parallel circuit, so, u = U1 = U2 = 3vr1 = U1 / I1 = 3 / 0.6 = 5 (Ω) and because, 1 / r = 1 / R1 + 1 / r2 = 1 / 5 + 1 / 10 = 3 / 10R = 10 / 3 (Ω) trunk current: I = u / r = 3 / (10 / 3) = 0.9 (a) answer: slightly. Hope to help you, if you have any questions, you can ask ~
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