Model Answer

From The Diagram

From Node b we get:

${\displaystyle V_{b}=-V_{1}=-15V}$

From Node d we get:

${\displaystyle V_{d}=V_{2}=-7V}$

It is clear that we must solve V_c, in order to complete Voltage definitions at all nodes. V_c will be found by applying KCL at Node c and solving resulting equations Follows:

${\displaystyle i_{3}=i_{1}+i_{2}}$

${\displaystyle {\frac {V_{c}}{R_{3}}}={\frac {V_{b}-V_{c}}{R_{1}}}+{\frac {V_{d}-V_{c}}{R_{2}}}}$

We can group like terms to get the following equation:

${\displaystyle V_{c}({\frac {1}{R_{1}}}+{\frac {1}{R_{2}}}+{\frac {1}{R_{3}}})={\frac {V_{b}}{R_{1}}}+{\frac {V_{d}}{R_{2}}}}$

Substitute values into previous equations you get:

${\displaystyle V_{c}({\frac {1}{20\Omega }}+{\frac {1}{5\Omega }}+{\frac {1}{10\Omega }})={\frac {-15V}{20\Omega }}+{\frac {-7V}{5\Omega }}}$

${\displaystyle V_{c}(0.35)=-2.15}$   thus   ${\displaystyle V_{c}=-6.14V}$

Thus now we can calculate Current through ${\displaystyle R_{3}}$ as follows:
${\displaystyle {\begin{matrix}\ I_{R3}&=&{\frac {V_{c}}{R_{3}}}\\\ \\\ &=&{\frac {-6.14}{10}}\\\ \\\ &=&-0.614A\end{matrix}}}$.

Thus the effective current through ${\displaystyle R_{3}}$ is in opposite direction to ${\displaystyle i_{3}}$ Just as we expected!