- #1

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We know that

- V

_{M}=V

_{Cl}

- R

_{1}=1000.0 Ω

- I

_{M}=0

- V

_{Na}=61 mV

- V

_{K}=-88 mV

- V

_{Cl}=-71 mV

Question is: what is R

_{2}?

I've written according to Kirchhoff's second law that

(1) V

_{M}+ V

_{Na}- R

_{1}I

_{Na}= 0

and (2) V

_{M}- V

_{K}- R

_{2}I

_{K}= 0.

We can also calculate that I

_{Cl}=0.

With these equations and according to Kirchhoff's first law we can solve (using absolute values) that R

_{2}=128.789 Ω that is a correct answer. The problem is that I wonder why V

_{M}is always positive? Shouldn't it be negative in the first equation? If we go through capacitor and then Na-part of the circuit, the capacitor will have a negative charge at the bottom and a positive charge at the top because so has V

_{Na}. (Instead, if we continue to K or Cl part of the circuit, the capacitor will have a positive charge at the bottom and a negative at the top.) So the first equation would be -V

_{M}+ V

_{Na}- R

_{1}I

_{Na}= 0 but then we get an incorrect answer. I see that I must have misunderstood something but I don't know what is it.

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