I have problems solving an electric circuit. I need to find the red drawn "I" (Current). However the way the electric circuit is drawn strikes me. I have very basic knowledge of physics and my skills are limited. The way I tried to attempt the problem is to look at the bottom left triangle and the top right triangle. I then used the Mesh-Current-Method for the bottom left triangle which I defined as: $m_1 = -U_q_1 + U_1 + U_3 + U_q_3$ ($U_1$ is the Voltage at $R_1$ and $U_3$ the Voltage at $R_3$).
This term must be equal 0 due to Kirchoff law. Knowing $U_q_1$ and $U_q_3$ (both 5V) we are left with $m_1 = U_1 + U_3$. As far as I know $U_1$ and $U_3$ cannot be negative, so I assumed that both are 0V (due to the addition). Repeating this for the upper triangle returns me $U_2 = 0$. With Ohm's law $I$ must be 0. However my own intuition is that I did something wrong. Any ideas?
Best Answer
As a general rule, clearly define all voltage (including polarities). This can be done by labeling either each component of the schematic with the a polarity (+ or $\rightarrow$). Alternatively, you can define a 'reference' for each node in the schematic (eg: by a letter), as shown below.
We will indicate the voltage of $A$ with respect to $B$ as $V_{AB}$.
For example, $U_{q1}=V_{FD}=-V_{DF}$
Now, applying Kirchhoff's Voltage Law (KVL) around loop A-E-F-D-A, we get:
$V_{AE}+V_{EF}+V_{FD}+V_{DA}=0$
ie:
$I_{3}R_{3}-U_{q3}+U_{q1}-I_{1}R_{1}=0$
Substituting known value, we get
$I_{1}=I_{3}$
Applying Kirchhoff's Current Law (KCL) at node A,
$I_{1}+I_{3}-I_{4}=0$
Which gives,
$I_{4}=2I_{1}=2I_{3}$, or $I_{3}=I_{4}/2$
Finally, applying KVL around loop A-B-C-F-E-A, we get:
$V_{AB}+V_{BC}+V_{CF}+V_{FE}+V_{EA}=0$
ie
$-I_{4}R_{2}+U_{q2}-I_{4}R_{4}+U_{q3}-I_{3}R_{3}=0$
Substituting known values,
$-I_{4}+5-I_{4}+5-I_{4}/2=0$
ie
$I_{4}=4 A$
so $I = -I_{4} = -4 A$