In the circuit shown,the currents (in A) thro | vedclass

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(C) Let the potential at the left node be $V_1 = 50\; V$ and at the right node be $V_2 = 0\; V$. However,it is easier to use nodal analysis. Let the p

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$4.5$ and $1$

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(C) Let the potential at the left node be $V_1 = 50\; V$ and at the right node be $V_2 = 0\; V$. However,it is easier to use nodal analysis. Let the potential at the left junction be $V_A$ and at the right junction be $V_B$.Applying Kirchhoff's Current Law $(KCL)$ at node $A$:$\frac{V_A - 50}{0} + \frac{V_A - V_B}{5 + 5} + \frac{V_A}{20} = 0$ (This is not ideal). Let's use the node potential method. Let the potential of the bottom wire be $0\; V$. The potential at the top left is $50\; V$ and top right is $30\; V$. Let the potential at the junction after the $50\; V$ battery be $V_1$ and after the $30\; V$ battery be $V_2$.Actually,using Millman's Theorem or nodal analysis: Let the potential at the left junction be $V_L$ and right junction be $V_R$.For the left branch: $(V_L - 50) / 0 = 0$ (ideal battery). Let's simplify: The circuit has two loops. Let $I_1$ be the current from the $50\; V$ battery and $I_2$ be the current from the $30\; V$ battery.Using nodal analysis,let the potential at the left vertical wire be $V_1$ and right be $V_2$. $V_1 = 50\; V$,$V_2 = 30\; V$.The resistance between them is $R_{eq} = 5 + 5 = 10\; \Omega$.The current flowing from left to right is $I = (50 - 30) / 10 = 2\; A$.At the left node,current $I_1$ leaves the battery. It splits into $2\; A$ (towards the right) and $I_{20} = 50 / 20 = 2.5\; A$ (downwards).So,$I_1 = 2 + 2.5 = 4.5\; A$.At the right node,current $I_2$ enters from the battery. The current $2\; A$ arrives from the left,and $I_{10} = 30 / 10 = 3\; A$ flows downwards.By $KCL$ at the right node: $I_2 + 2 = 3 \implies I_2 = 1\; A$.Wait,re-evaluating: The current through $50\; V$ battery is $4.5\; A$ and through $30\; V$ battery is $1\; A$.

In the circuit shown,the currents (in $A$) through the $50\; V$ and $30\; V$ batteries are,respectively:

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