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QUESTION:

The switch above the 12 V source in the circuit shown below has been closed since just after the wheel was invented. It is finally thrown open at t = 0.

(a) Compute the circuit time constant (in ms).

(b) Obtain an expression for v(t) valid for t > 0 (you can use 'exp' and ^ symbols) (here X is a variable)

I have to compute v(t) as given in the circuit but I have doubt as I cant find out across which two nodes the potential difference is v(t).

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    \$\begingroup\$ We can't either, as there is no reference point. I can guess that the reference is probably where the - of the 12V source is connected. This is a poorly formed exercise, unless there is some extra information that you have not posted. \$\endgroup\$
    – Eugene Sh.
    Commented Oct 8, 2020 at 14:23

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The *circuit has a time constant of R.C The switch is perhaps there to mislead you.

Since it's homework I'll leave you to calculate R. What's the value of X btw ? You may need to include that in your answer.

V(t) is easily calculated if you've been following your lectures. It'll help somewhat if you simplify the circuit using Norton's or Thevenin's theorem.

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  • \$\begingroup\$ I think you missed the point of the question. There is no clearly indicated reference point for v(t), so it can not be "easily calculated". \$\endgroup\$
    – Elliot Alderson
    Commented Oct 8, 2020 at 20:28
  • \$\begingroup\$ I think you need to infer B- as gnd/common. \$\endgroup\$
    – Graham Stevenson
    Commented Oct 8, 2020 at 21:26
  • \$\begingroup\$ Sorry, I don't see B- marked on the schematic. And any inference here seems dangerous and unwarranted to me. \$\endgroup\$
    – Elliot Alderson
    Commented Oct 8, 2020 at 22:08
  • \$\begingroup\$ As in the battery negative connection (as marked). In this case, either one makes an intelligent deduction or there's no answer ! As noted in another comment 'The question is bad, and strictly speaking contains insufficient information for anyone to be able to solve it'. Whilst I agree, it seems not too difficult to deduce what was meant. \$\endgroup\$
    – Graham Stevenson
    Commented Oct 9, 2020 at 0:33

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