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Linear Circuit

Is this solution of linear circuit correct ? Have I applied KCL and KVL correctly?

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    \$\begingroup\$ Since I see an equation for V1 but V1 isn't marked on the circuit, and I see no explicit value for R1, I'm going to suggest no. \$\endgroup\$ – Brian Drummond Feb 14 '16 at 13:00
  • \$\begingroup\$ @BrianDrummond V1 is the potential drop across R1 \$\endgroup\$ – Maya Feb 14 '16 at 15:47
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Since there's 1.5mA through R1, there also has to be 1.5mA through the device in series with it

Since there's 2 volts aceoss R2, there also has to be 2v across the device in parallel with it.

Since there's 2 volts across R2, 4 volts across the device in series with R1 and Vs is 8 volts, then there has to be 2 volts across R1.

With all that in mind, we can say:

enter image description here

and the proof is in the voltage divider.

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No, because Vs and R2 are not in parallel. However, device B's voltage drop and R2 are in parallel.

P.S. Simpler KVL than [Vs, Device A, R1, R2] can be [Vs, Device A, R1, Device B]

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  • \$\begingroup\$ I don't understand. Why are Vs and R2 not in parallel? \$\endgroup\$ – Maya Feb 14 '16 at 15:53
  • \$\begingroup\$ Because in order for two components to be in parallel they must connect with two nodes directly. Vs shares two nodes with the whole circuit (device a + R1 + (R2 in parallel with device b). A perfect example of two components sharing two nodes directly are Device b and R2, thus in parallel. \$\endgroup\$ – Iancovici Feb 14 '16 at 15:56
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It seems you determined that R2 is 4 kΩ. That's right. It doesn't seem that you solved for R1 though. That's also very easy.

You can go thru the general equations and solve things that way, but it's actually very easy to solve for both resistors by inspection and then one divide on a calculator.

Hint: You can see from inspection the current thru and the voltage across R1. Likewise for R2.

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  • \$\begingroup\$ how do you get from 8/.5 in the question to 4? \$\endgroup\$ – Brian Drummond Feb 14 '16 at 16:22
  • \$\begingroup\$ @Brian: Oops. Somehow I was thinking 8 * .5. \$\endgroup\$ – Olin Lathrop Feb 14 '16 at 20:17

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