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I've been having difficulty in understanding why the voltage across the resistor $3$ is $10V$. My main problem is understanding why the $12V$ resistor couldnt for example drop $1V$ on the first resistor and then drop $11V$ on $R_3$. Could anyone help explain this?

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The voltage on $R_3$ is 10 V only in the case where $R_1 = 0$, in this case there is no drop on $R_1$ and the resistor $R_3$ is in parallel with the source $V_1 = 10$ V, so that the voltage on $R_3$ is exactly $V_1$.

If $R_1 \not = 0$, you have to solve the circuit to find $I_3$. There are many ways of doing this, Kirchhoff's circuit laws are in general easy to understand and use. When you apply them you get

\begin{eqnarray} I_1R_1 + 30I_3 &=& 10 \\ 10I_2 + 30I_3 &=& 12 \\ I_1 + I_2 - I_3 &=& 0 \end{eqnarray}

Whose solution is

\begin{eqnarray} I_1 &=& \frac{40}{40 R_1+300}, \\ I_2 &=& \frac{12 (R_1+30)}{40 R_1+300}-\frac{300}{40 R_1+300}\\ I_3 &=& \frac{12 R_1}{40 R_1+300}+\frac{100}{40 R_1+300} \end{eqnarray}

Note that the voltage accross $R_3$ is $I_3R_3$ and depends on the value of $R_1$

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  • $\begingroup$ I'm sorry, I don't think I was very clear in my question. The first part of your answer is The situation in which my question was addressed. When R1 is 0. Do you follow? $\endgroup$
    – Jake Rose
    May 9 '17 at 0:29
  • $\begingroup$ @JakeRose Is then your question solved? $\endgroup$
    – caverac
    May 9 '17 at 0:33
  • $\begingroup$ Maybe I'm being silly and this is quite a trivial problem but could you explain why the 12v supply is limited by the 10v supply? $\endgroup$
    – Jake Rose
    May 9 '17 at 18:22
  • $\begingroup$ @JakeRose It is limited by the resistor $R_2$ $\endgroup$
    – caverac
    May 9 '17 at 19:12
  • $\begingroup$ How do we know there must be 2v on R2? $\endgroup$
    – Jake Rose
    May 9 '17 at 19:33

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