Find the current in the 10,Omega resistance. | vedclass

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(C) Let the potential at the bottom node be $0\,V$ and the potential at the top node be $V$. Using Kirchhoff's Current Law $(KCL)$ at the top node:$\f

Vedclass · Accepted Answer

$0.45\, A$ from $P_2$ to $P_1$

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(C) Let the potential at the bottom node be $0\,V$ and the potential at the top node be $V$. Using Kirchhoff's Current Law $(KCL)$ at the top node:$\frac{V - 5}{2} + \frac{V}{10} + \frac{V - 5}{2} = 0$Multiply by $10$ to simplify:$5(V - 5) + V + 5(V - 5) = 0$$5V - 25 + V + 5V - 25 = 0$$11V = 50$$V = \frac{50}{11} \approx 4.545\,V$The current $I$ through the $10\,\Omega$ resistor flows from the top node $(P_2)$ to the bottom node $(P_1)$:$I = \frac{V}{10} = \frac{50/11}{10} = \frac{5}{11} \approx 0.4545\,A$Rounding to two decimal places,the current is $0.45\,A$ from $P_2$ to $P_1$.

Find the current in the $10\,\Omega$ resistance. The circuit consists of two $5\,V$ batteries with internal resistances of $2\,\Omega$ each,connected in parallel with a $10\,\Omega$ resistor.

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