0
\$\begingroup\$

In case of parallel connection, is the statement "Depending on the current(electric current) value, logic (0 or 1) is determined." True or False? I do not understand why current cannot affect the voltage value in this case. Don't we need another assumption about the resistors having the same value in order for the statement to be false? I am confused.

This is a TF question from my logic design class. The answer provided is false because it is determined by voltage. I marked this as true because I thought current can affect the voltage therefore determine the logic value. But when I mailed my professor he said this is not true in case parallel connection. I am wondering why this is the case.

\$\endgroup\$
  • \$\begingroup\$ what do you mean by current value? .... does it mean the most recent value, or does it mean the value of electric current ? .... please clarify the question .... do not add additional info into comments \$\endgroup\$ – jsotola Apr 23 '18 at 16:17
  • \$\begingroup\$ Many logic circuits have input impedance high enough that it becomes difficult to even measure the input current. I'm not sure what parallel circuits has to do with it. \$\endgroup\$ – The Photon Apr 23 '18 at 16:25
  • \$\begingroup\$ zero or one can represent any state, you only need some way to discriminate between the two. You could develop a system to measure current an not voltage, it would be more difficult. \$\endgroup\$ – Voltage Spike Apr 23 '18 at 16:50
1
\$\begingroup\$

It's not a very clear question, but on logic ICs, it's the input voltage that determines if that input is high or low. The IC doesn't know or care what current is flowing from the previous IC's output.

schematic

simulate this circuit – Schematic created using CircuitLab

So in the little diagram above, pressing SW2 changes the current flowing out of BUF1, but BUF2 doesn't care.

\$\endgroup\$
  • \$\begingroup\$ Thank you. I can see how the current is irrelevant. \$\endgroup\$ – Seung Joon Park Apr 23 '18 at 17:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.