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- Homework Statement
- Consider the circuit section of the figure. Determine the potential difference ##V_a-V_b##.

Data: ##\varepsilon_1=6\, \textrm{V}##; ##\varepsilon_2=7\, \textrm{V}##; ##R_1=2\, \textrm{k}\Omega##; ##R_2=4\, \textrm{k}\Omega##.

Sol.: ##11\, \textrm{V}##

- Relevant Equations
- Ohm's law

The figure is:

I have the solution to this problem:

We have two distinct branches

$$V_a-V_b=\overbrace{(V_a-V_c)}^{\textrm{INI}-\textrm{FIN}}+\overbrace{(V_c-V_b)}^{\textrm{FIN}-\textrm{INI}}$$

They have different intensities: ##3\, \textrm{mA}## and ##2\, \textrm{mA}##

##V_A-V_C\rightarrow## We have battery and resistor

$$V_A-V_C=+\overbrace{R_1I_1}^{V_1}+\varepsilon_1$$

##V_C-V_B\rightarrow## High potential ##\rightarrow V_B##

$$V_C-V_B=-\overbrace{R_2I_2}^{V_2}+\varepsilon_2$$The question that I have is: If they tell me the potential ##V_a-V_b## the green arrow it should not go the other way. Like the blue one I have drawn. So I would make ##V_a-V_b=(V_c-V_b)+(V_a-V_c)##.

I have the solution to this problem:

We have two distinct branches

$$V_a-V_b=\overbrace{(V_a-V_c)}^{\textrm{INI}-\textrm{FIN}}+\overbrace{(V_c-V_b)}^{\textrm{FIN}-\textrm{INI}}$$

They have different intensities: ##3\, \textrm{mA}## and ##2\, \textrm{mA}##

##V_A-V_C\rightarrow## We have battery and resistor

$$V_A-V_C=+\overbrace{R_1I_1}^{V_1}+\varepsilon_1$$

##V_C-V_B\rightarrow## High potential ##\rightarrow V_B##

$$V_C-V_B=-\overbrace{R_2I_2}^{V_2}+\varepsilon_2$$The question that I have is: If they tell me the potential ##V_a-V_b## the green arrow it should not go the other way. Like the blue one I have drawn. So I would make ##V_a-V_b=(V_c-V_b)+(V_a-V_c)##.