Process Engineering Division – Instrumentation and Automation Section – Daily Topics – Topic No. 2020-12-11
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This post was last edited by It will be better tomorrow on 2020-12-14 08:17 in the Process Engineering Section – Instrumentation and Automation Forum – Daily Topics – Topic No. 2020-12-11. Discussion topic: 323. Given a 24VDC power supply unit used to power several 12V indicator lights in series, how many indicator lights can be connected in series if they are to function properly? If each indicator light has a resistance of 5 ohms, what is the current flowing through each one? (5 points are given only to those who provide the correct answer; additional points are awarded for a correct calculation process or detailed explanation.)
In series, the voltage is divided as 24/12=2; thus 2 lights can be connected in series. The current is I=U/R=24/(5×2)=2.4A, so the current flowing through each indicator light is 2.4A
In series, the voltage is divided as 24/12=2; thus 2 lights can be connected in series. The current is I=U/R=12/5=2.4A, so the current flowing through each indicator light is 2.4A
The voltages in series are added together: 24V = 12V + 12V, which provides power for 2 12V indicator lights connected in series； The current in the circuit is I=24/(5+5)=2.4A, and the current flowing through each indicator light is 2.4A
The rated current for each indicator light is 12/5 = 2.4 (A). After connecting resistors in series, the resistance becomes 24/2.4 = 10 (ohms). By adding the resistances of the series connection, it results in two indicator lights connected in series. The current passing through each indicator light is 2.4A
Two in series: 24/12=2, current is 2.4A; 24/(5*2)=2.4
In series, the voltage is divided as 24/12=2; thus 2 lights can be connected in series. The current is I=U/R=24/(5×2)=2.4A, so the current flowing through each indicator light is 2.4A
In series, the voltage is divided as 24/12=2; thus 2 lights can be connected in series. The current is I=U/R=12/5=2.4A, so the current flowing through each indicator light is 2.4A
It is known that a 24VDC power supply unit is used to power several 12V indicator lights in series; if the indicators work properly, 2 indicator lights can be connected in series: 24/12=2. If each indicator light has a resistance of 5 ohms, and the current flowing through each indicator light is 2.4 A: I = U/R = 24/(5*2) = 2.4 A.
In series, the voltage is divided as 24/12=2; thus 2 lights can be connected in series. The current is I=U/R=24/(5×2)=2.4A, so the current flowing through each indicator light is 2.4A
In series, the voltage is divided as 24/12=2; thus 2 lights can be connected in series. The current is I=U/R=24/(5×2)=2.4A, so the current flowing through each indicator light is 2.4A