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Just a general question about voltage graphs for first order circuits.

For example in this circuit:

enter image description here

The graph of v(t), the voltage across the capacitor may turn out to be something like this (values not correct)

enter image description here

Would the voltage across the 6k resistor be the opposite of this graph? For example if the values in the graph were correct would it start at -9V?

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  • \$\begingroup\$ The voltage across the 6k resistor just before the switch is open equal to 12*6/(4+6+2) = 6V, so the graph should start at 6V \$\endgroup\$
    – G36
    Commented Apr 10, 2016 at 15:31
  • \$\begingroup\$ Well yes the values in the graph are not correct, I was just using it as an example. I was saying that if those values were correct for the voltage across the capacitor, would the values for the resistor be the negative of it? \$\endgroup\$
    – David M
    Commented Apr 10, 2016 at 18:32

2 Answers 2

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For questions like this, we prefer not to give complete solutions. Instead, we'd rather guide you to a solution so that you learn how to solve the problem yourself. So here's a couple of hints.

Hint: In the DC steady-state solution before the switch is opened, what is the relationship between \$v(t)\$ and \$v_1(t)\$?

Hint to the hint: In DC steady state, what is the current through the upper-right 2 kOhm resistor?

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No, the voltage v1(t) across the 6K resistors in not the opposite of the graph, assuming that is the actual solution of v(t).

For t > 0, assuming you have v(t) already, you have to solve for v1(t) with the applicable circuit using v(t). In this case, the applicable circuit is just the 2K and 6K resistors and the capacitor in series. The given v(t) is applied as the voltage across the capacitor.

By the way, if you observe the signs of the definition of v1(t) on the schematics, v1(t) never goes negative as given.


Responding to the comment: If I define a current going clockwise on the circuit for t>0, I can start off with: $$ V_C + V_{R6K} + V_{R2K} = 0 $$ Then observing the definitions on the schematic: $$ V_C = v(t) $$ But $$ V_{R6K} = -v_1(t) $$ Therefore $$ v(t) - v_1(t) + V_{R2K} = 0 $$

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  • \$\begingroup\$ Okay, thanks. Where I'm confused is wouldn't the voltage across the resistor have to be negative if v(t) was positive at t > 0? In the loop wouldn't it be Vc + Vr = 0? \$\endgroup\$
    – David M
    Commented Apr 11, 2016 at 0:42

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