In the circuit shown in the figure below,whic | vedclass

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(B) $1$. The circuit contains two batteries of $10\,V$ and $20\,V$ in series, and resistors of $7.5\,\Omega$, $0.5\,\Omega$, $1\,\Omega$, and $1\,\Ome

Vedclass · Accepted Answer

The potential at $Q$ is $-1\,V$

Vedclass · Answer

(B) $1$. The circuit contains two batteries of $10\,V$ and $20\,V$ in series, and resistors of $7.5\,\Omega$, $0.5\,\Omega$, $1\,\Omega$, and $1\,\Omega$.$2$. The net electromotive force $(EMF)$ is $E_{net} = 20\,V - 10\,V = 10\,V$. The $20\,V$ battery dominates, so the current flows clockwise.$3$. The total resistance of the circuit is $R_{total} = 7.5\,\Omega + 0.5\,\Omega + 1\,\Omega + 1\,\Omega = 10\,\Omega$.$4$. The current in the circuit is $I = E_{net} / R_{total} = 10\,V / 10\,\Omega = 1\,A$ (clockwise).$5$. Point $G$ is grounded, so $V_G = 0\,V$. Since $S$ and $R$ are connected to $G$ by ideal wires, $V_S = 0\,V$ and $V_R = 0\,V$.$6$. Moving from $R$ to $Q$ through the $1\,\Omega$ resistor in the direction opposite to the current, the potential increases: $V_Q = V_R + I 	imes 1\,\Omega = 0 + 1\,A 	imes 1\,\Omega = 1\,V$.$7$. Moving from $S$ to $P$ through the $7.5\,\Omega$ resistor in the direction of the current, the potential decreases: $V_P = V_S - I 	imes 7.5\,\Omega = 0 - 1\,A 	imes 7.5\,\Omega = -7.5\,V$.$8$. Comparing these results with the options, statement $A$ is correct $(V_P = -7.5\,V)$, statement $B$ is incorrect $(V_Q = 1\,V$, not $-1\,V)$, statement $C$ is correct $(V_R = 0\,V)$, and statement $D$ is correct $(V_S = 0\,V)$.$9$. Therefore, the incorrect statement is $B$.

In the circuit shown in the figure below,which of the following statements is incorrect?

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