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enter image description here

Hi all, So the question for this circuit was: Using logic gates design a circuit which turns on a light if any one of a car's four doors are opened. Assume that each door is equipped with a switch which is open-circuit if the door is opened.

I do not understand why when a switch is closed, the current flows to ground. Can anyone explain this?

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  • \$\begingroup\$ Because one side of the switch is connected to ground. I'm not sure what else to say about this. Maybe if you keep explaining your reasoning, we'll be able to spot what you're misunderstanding. \$\endgroup\$ – The Photon Jun 3 '16 at 17:05
  • \$\begingroup\$ Do you understand that if S1 is cloed, then the node that connects R1, S1, and R5 will be connected directly to ground? \$\endgroup\$ – The Photon Jun 3 '16 at 17:05
  • \$\begingroup\$ Have you studied ohms law? \$\endgroup\$ – Andy aka Jun 3 '16 at 17:06
  • \$\begingroup\$ Many people confuse "close" and "open" with plumbing. If you close a faucet, water stops flowing. If you close a switch, you allow electrical current to flow. Similarly, if you open a faucet, water will flow, but if you open a switch, no current can flow. \$\endgroup\$ – David Drysdale Jun 3 '16 at 17:36
  • \$\begingroup\$ A closed switch behaves as a wire. So when S1 is closed the bottom end of R1 is connected directly to ground. So what is the voltage at the bottom end of R1? ("Bottom" as in the end that's closer to the bottom edge of the circuit diagram) \$\endgroup\$ – user253751 Jun 4 '16 at 10:38
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Under KVL's hypothesis which is quasi-static and conversative system. The current go from the higher potential to the lower potential. In this case, when you close the switch, there is no current going into the base because the emitter is grounded therefore there is no current in the base and the light won't light. Here the circuit when the switch is close:

schematic

simulate this circuit – Schematic created using CircuitLab

It is pretty much self-explanatory with the image.

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  • \$\begingroup\$ @Chora if your question is answered, would like to click the button next the question to make the question "answered" sure the question isn't anymore in the question feed. \$\endgroup\$ – MathieuL Jun 3 '16 at 17:31
  • \$\begingroup\$ But if any of the switches say S1 is closed, this does not mean that the base current is zero as the case in your image !, in the circuit diagram that Chora posted, if any of the switches is closed,the base current decreases but if all of the 4 switches are closed the base current is zero \$\endgroup\$ – Elbehery Jun 3 '16 at 17:37
  • \$\begingroup\$ Read the question "I do not understand why when a switch is closed, the current flows to ground. Can anyone explain this?" It is just about 1 circuit \$\endgroup\$ – MathieuL Jun 3 '16 at 17:39
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enter image description here

You can think of it from studying physics of semiconductor of the transistor used in this circuit. When you leave the switch open, there will be a current in the base; thus, the base junction will be high. What I mean by high is higher than the collector voltage and the emitter voltage as well. When the switch is pressed (as shown in the left image), the base voltage will be grounded so it will not be higher than emitter voltage.

Remember the case of a transistor acting as a switch mode (as shown in the right image) the base voltage should be higher than both collector and emitter voltage.

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