Find the electric potential at points C and D | vedclass

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(B) The point $P$ is grounded,so its potential $V_P = 0\,V$.The total electromotive force $(EMF)$ in the circuit is $E_{net} = 12\,V - 6\,V = 6\,V$.Th

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$V_C = 9\,V$; $V_D = 6\,V$

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(B) The point $P$ is grounded,so its potential $V_P = 0\,V$.The total electromotive force $(EMF)$ in the circuit is $E_{net} = 12\,V - 6\,V = 6\,V$.The total resistance in the circuit is $R_{eq} = 1\,\Omega + 2\,\Omega + 3\,\Omega = 6\,\Omega$.The current $I$ flowing in the circuit is $I = \frac{E_{net}}{R_{eq}} = \frac{6\,V}{6\,\Omega} = 1\,A$ (flowing clockwise).Starting from point $P$ $(0\,V)$ and moving towards $C$ through the $12\,V$ battery and $1\,\Omega$ and $2\,\Omega$ resistors:$V_C = V_P + 12\,V - I(1\,\Omega + 2\,\Omega) = 0 + 12 - 1(3) = 9\,V$.Starting from point $P$ $(0\,V)$ and moving towards $D$ through the $6\,V$ battery (in reverse direction) and $3\,\Omega$ resistor:$V_D = V_P - 6\,V + I(3\,\Omega) = 0 - 6 + 1(3) = -3\,V$. Wait,re-evaluating the circuit: The current flows from the $12\,V$ source through the resistors to the $6\,V$ source. $V_C = 12 - 1(1+2) = 9\,V$.$V_D = V_C - I(3) = 9 - 1(3) = 6\,V$.Thus,$V_C = 9\,V$ and $V_D = 6\,V$.

Find the electric potential at points $C$ and $D$ in the given circuit.

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