In the given circuit,the potentials at points | vedclass

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(B) The circuit consists of a $12\ V$ battery and a $6\ V$ battery in series,with the junction between them grounded $(0\ V)$.The total electromotive

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$11\ V, 9\ V, 6\ V$

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(B) The circuit consists of a $12\ V$ battery and a $6\ V$ battery in series,with the junction between them grounded $(0\ V)$.The total electromotive force $(EMF)$ in the loop is $12\ V - 6\ V = 6\ V$.The total resistance in the circuit is $R_{eq} = 1\ \Omega + 2\ \Omega + 3\ \Omega = 6\ \Omega$.Using Ohm's law,the current $i$ in the circuit is $i = \frac{V_{net}}{R_{eq}} = \frac{6\ V}{6\ \Omega} = 1\ A$.The potential at point $A$ is $12\ V$ (relative to the ground).Potential at $B$: $V_B = V_A - iR_1 = 12\ V - (1\ A 	imes 1\ \Omega) = 11\ V$.Potential at $C$: $V_C = V_B - iR_2 = 11\ V - (1\ A 	imes 2\ \Omega) = 9\ V$.Potential at $D$: $V_D = V_C - iR_3 = 9\ V - (1\ A 	imes 3\ \Omega) = 6\ V$.Wait,let's re-evaluate the circuit. The bottom of the $6\ V$ battery is connected to $D$. Since the junction between the batteries is $0\ V$,and the $6\ V$ battery is connected such that its negative terminal is at $D$,the potential at $D$ is $-6\ V$. Let's re-check the loop: $12 - i(1+2+3) - 6 = 0 \implies 6 = 6i \implies i = 1\ A$. Starting from $A$ $(12\ V)$: $V_B = 12 - 1(1) = 11\ V$,$V_C = 11 - 1(2) = 9\ V$,$V_D = 9 - 1(3) = 6\ V$. The potential at $D$ is $6\ V$ relative to the ground.

In the given circuit,the potentials at points $B$,$C$,and $D$ will be:

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