Subjective Type

Find the magnitude and direction of the current flowing through the resistance $$R$$ in the circuit shown in Fig. if the emf's of the sources are equal to $$\varepsilon_{1} = 1.5\ V$$ and $$\varepsilon_{2} = 3.7\ V$$ and the resistances are equal to $$R_{1} = 10\Omega, R_{2} = 20\Omega, R = 5.0\Omega$$. The internal resistances of the sources are negligible.
Solution
Make the current distribution, as shown in the diagram.
Now, in the loop $$12341$$, applying $$-\triangle \varphi = 0$$
$$iR + i_{1}R_{1} + \xi_{1} = 0 (1)$$
and in the loop $$23562$$
$$iR - \xi_{2} + (i - i_{1})R_{2} = 0 (2)$$
Solving (1) and (2), we obtain current through the resistance $$R$$,
$$i = \dfrac {(\xi_{2}R_{1} - \xi_{1}R_{2})}{RR_{1} + RR_{2} + R_{1}R_{2}} = 0.02\ A$$
and it is directed from left to the right.
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