For the resistance network shown in the figur | vedclass

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(B) The circuit can be simplified by recognizing symmetry. The upper branch consists of three $2 \ \Omega$ resistors in series,giving a total resistan

Vedclass · Accepted Answer

$(A, B, D)$

Vedclass · Answer

(B) The circuit can be simplified by recognizing symmetry. The upper branch consists of three $2 \ \Omega$ resistors in series,giving a total resistance of $R_1 = 2+2+2 = 6 \ \Omega$. The lower branch consists of three $4 \ \Omega$ resistors in series,giving a total resistance of $R_2 = 4+4+4 = 12 \ \Omega$.These two branches are in parallel across the $12 \ V$ source.Equivalent resistance $R_{eq} = \frac{R_1 R_2}{R_1 + R_2} = \frac{6 	imes 12}{6 + 12} = \frac{72}{18} = 4 \ \Omega$.Total current $I_1 = \frac{V}{R_{eq}} = \frac{12}{4} = 3 \ A$.Current in the upper branch $I_2 = \frac{V}{R_1} = \frac{12}{6} = 2 \ A$.Since the potential drops across the first $2 \ \Omega$ resistor and the first $4 \ \Omega$ resistor are proportional to their resistances,the potentials at nodes $P$ and $Q$ are equal $(V_P = V_Q)$. Thus,the current through the $1 \ \Omega$ resistor connecting $P$ and $Q$ is zero.Similarly,$V_S = V_T$,so the current through the $1 \ \Omega$ resistor connecting $S$ and $T$ is zero.Since $V_P = V_Q$ and $V_S = V_T$,and the potential decreases along the branches,the potential at $S$ is less than the potential at $Q$ because $S$ is further along the path from the source than $Q$ is relative to the potential drop.Therefore,options $(A)$,$(B)$,and $(D)$ are correct.

For the resistance network shown in the figure,choose the correct option$(s)$.

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