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(B) Let the potential at the junction $O$ be $V$. Applying Kirchhoff's Current Law $(	ext{KCL})$ at junction $O$:$i_1 = i_2 + i_3$Using Ohm's law,the

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$0.2$

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(B) Let the potential at the junction $O$ be $V$. Applying Kirchhoff's Current Law $(	ext{KCL})$ at junction $O$:$i_1 = i_2 + i_3$Using Ohm's law,the currents can be expressed as:$\frac{10 - V}{10} = \frac{V - 6}{20} + \frac{V - 5}{30}$To solve for $V$,multiply the entire equation by the least common multiple of the denominators,which is $60$:$6(10 - V) = 3(V - 6) + 2(V - 5)$$60 - 6V = 3V - 18 + 2V - 10$$60 - 6V = 5V - 28$$11V = 88$$V = 8 	ext{ V}$Now,calculate the current $i_1$ flowing through $R_1$:$i_1 = \frac{10 - V}{10} = \frac{10 - 8}{10} = \frac{2}{10} = 0.2 	ext{ A}$

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