In the figure shown,the current in the 10 , V | vedclass

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(C) To find the current in the $10 \, V$ battery,we can simplify the circuit using Thevenin's theorem or nodal analysis. Let the node between the $5 \

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$0.21 \, A$ from positive to negative terminal.

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(C) To find the current in the $10 \, V$ battery,we can simplify the circuit using Thevenin's theorem or nodal analysis. Let the node between the $5 \, \Omega$ and $10 \, \Omega$ resistors be $A$ and the bottom node be $B$. Applying Kirchhoff's Current Law at the node above the $10 \, \Omega$ resistor,let the potential be $V$. $\frac{V - 20}{5 + 2} + \frac{V}{10} + \frac{V - 10}{4} = 0$ $\frac{V - 20}{7} + \frac{V}{10} + \frac{V - 10}{4} = 0$ Multiplying by $140$ ($LCM$ of $7, 10, 4$): $20(V - 20) + 14V + 35(V - 10) = 0$ $20V - 400 + 14V + 35V - 350 = 0$ $69V = 750$ $V = \frac{750}{69} \approx 10.87 \, V$ The current $I$ through the $10 \, V$ battery is $I = \frac{V - 10}{4} = \frac{10.87 - 10}{4} = \frac{0.87}{4} = 0.2175 \, A$. Since the potential $V$ is greater than $10 \, V$,the current flows from the positive terminal to the negative terminal.

In the figure shown,the current in the $10 \, V$ battery is close to

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