# Voltage at node A respective to the ground

I want to calculate the voltage at node $$A$$ respective to the ground. $$R_1=1.8\text{k}\Omega, R_2=3.8\text{k}\Omega, R_3=1.8\text{k}\Omega, R_4=5.8\text{k}\Omega, R_5=2.4\text{k}\Omega$$ 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.

• Are you sure that $2.64$ is the right answer because online review system says it's not . I don't know why Oct 30, 2019 at 17:37
• 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. Oct 31, 2019 at 10:02
• Answer updated. You can check the results at circuitlab.com. Oct 31, 2019 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:

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