I want to calculate the voltage at node $A$ respective to the ground.

enter image description here

$R_1=1.8\text{k}\Omega, R_2=3.8\text{k}\Omega, R_3=1.8\text{k}\Omega, R_4=5.8\text{k}\Omega, R_5=2.4\text{k}\Omega$ and $V1=4.3V, V2=2.3V$

I know that voltage $V=-V2+V1=2V$ then $R_1$ and $R_2$ in series so $R_{12}=R_1+R_2=5.6\text{k}\Omega$ then the current is $I=\frac{V}{R_{12}}=0.356\,\text{mA}$

But how do I calculate the voltage at node $A$ respective to the ground? Is it over $R_{45}$ or $R_{3}$?


If you consider the RHS loop of your circuit and using $i_3$ as the current going through $R_3$ (in the same direction as $i$), $i_4$ the current going through $R_4$ and $i_5$ the current going through $R_5$, you have (assuming your polarity of $V_1$ and $V_2$ is correct):

$$ -V_2 + V_1 + R_3 i_3 - V_A = 0\\ V_A = -R_4 i_4\\ V_A = -R_5 i_5\\ i_3 = i_4 + i_5 $$

which allows you to express $i_3$ as a function of $V_A$:

$$ i_3 = -V_A \left( \frac{1}{R_4} + \frac{1}{R_5} \right)$$

And you can use this in your loop equation to get $V_A$ as a function of $V_1$ and $V_2$:

$$ -V_2 + V_1 - R_3 V_A \left( \frac{1}{R_4} + \frac{1}{R_5} \right) - V_A= 0$$

which gives you $V_A$:

$$V_A = \frac{V_1 - V_2}{1 + R_3 \left( \frac{1}{R_4} + \frac{1}{R_5} \right)}$$

and if you plug the numerical values in you get V_A = 0.9707V.

  • $\begingroup$ Are you sure that $2.64$ is the right answer because online review system says it's not . I don't know why $\endgroup$ – engineerstudent Oct 30 '19 at 17:37
  • $\begingroup$ OK. I made a mistake assuming that the current going through $R_3$ is the same as the current going through $R_1$ and $R_2$. It isn't. Will update my answer. $\endgroup$ – am304 Oct 31 '19 at 10:02
  • $\begingroup$ Answer updated. You can check the results at circuitlab.com. $\endgroup$ – am304 Oct 31 '19 at 10:15

$R_4 \parallel R_5 $ and $R_3$ form a voltage divider:

$$V_A = \frac {R_4 \parallel R_5}{R_4 \parallel R_5 + R_3} \times (V_1 - V_2)$$

Or alternatively using series/parallel rules and Ohm's Law:

$$I_A = \frac {V_1 - V_2}{R_4 \parallel R_5 + R_3}$$ $$V_A = I_A \ (R_4 \parallel R_5)$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.