Find the current through the 2,Omega resistan | vedclass

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Question

(A) The circuit consists of two loops connected by a central $2\,\Omega$ resistor. Let the potential at the node between the $5\,\Omega$ resistor and

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$\frac{10}{17}$

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(A) The circuit consists of two loops connected by a central $2\,\Omega$ resistor. Let the potential at the node between the $5\,\Omega$ resistor and the $2\,\Omega$ resistor be $V_1$,and the potential at the node between the $10\,\Omega$ resistor and the $2\,\Omega$ resistor be $V_2$. Assuming the bottom wire is at $0\,V$ potential:Applying Kirchhoff's Current Law $(KCL)$ at node $V_1$:$\frac{V_1 - 10}{5} + \frac{V_1 - V_2}{2} = 0$$2(V_1 - 10) + 5(V_1 - V_2) = 0$$7V_1 - 5V_2 = 20$ --- (Equation $1$)Applying $KCL$ at node $V_2$:$\frac{V_2 - V_1}{2} + \frac{V_2 - 20}{10} = 0$$5(V_2 - V_1) + 1(V_2 - 20) = 0$$-5V_1 + 6V_2 = 20$ --- (Equation $2$)Solving the system of equations:Multiply (Equation $1$) by $6$ and (Equation $2$) by $5$:$42V_1 - 30V_2 = 120$$-25V_1 + 30V_2 = 100$Adding these gives $17V_1 = 220$,so $V_1 = \frac{220}{17}\,V$.Substituting $V_1$ into (Equation $2$):$6V_2 = 20 + 5(\frac{220}{17}) = \frac{340 + 1100}{17} = \frac{1440}{17}$$V_2 = \frac{240}{17}\,V$.The current $I$ through the $2\,\Omega$ resistor is $I = \frac{V_2 - V_1}{2} = \frac{1}{2} (\frac{240}{17} - \frac{220}{17}) = \frac{1}{2} (\frac{20}{17}) = \frac{10}{17}\,A$.

Find the current through the $2\,\Omega$ resistance in the given circuit. (in $A$)

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