In a simple series circuit, using Ohms law, with a known current of 5 amps, and a resistive load of 30 watts, what is the voltage drop?
回答 (4)
2017-09-13 10:23 am
Ohm's law: V = IR
Thus, P = I²R = I•IR = IV
Then, V = P/I
Voltage drop, V = P/I = 30/5 = 6 V
Thus, P = I²R = I•IR = IV
Then, V = P/I
Voltage drop, V = P/I = 30/5 = 6 V
2017-09-13 10:32 am
6v.
2017-09-15 1:28 pm
30W/5A=6V
2017-09-13 12:11 pm
(A^2)*(R) = 30W
R = (30/25)Ohms
Using Ohm's law:
(5A)*[(30/25)Ohms] = 6 Volts
R = (30/25)Ohms
Using Ohm's law:
(5A)*[(30/25)Ohms] = 6 Volts
收錄日期: 2021-04-24 00:41:03
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