Voltage at node A respective to the ground

Question

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 $R_{3}$?

Answer 1

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.

Answer 2

$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:

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