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(C) Before adding the dotted lines,the five resistors are connected in series. Therefore,the equivalent resistance $R_{before} = 1 \ \Omega + 1 \ \Ome

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$\frac{5}{3}$

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(C) Before adding the dotted lines,the five resistors are connected in series. Therefore,the equivalent resistance $R_{before} = 1 \ \Omega + 1 \ \Omega + 1 \ \Omega + 1 \ \Omega + 1 \ \Omega = 5 \ \Omega$.After adding the two dotted lines (which represent short-circuiting paths),the circuit configuration changes. The dotted lines connect nodes that are at the same potential or effectively bypass certain resistors. Specifically,the two dotted lines connect the top two nodes and the bottom two nodes respectively. This effectively puts the middle three resistors in parallel with the short-circuiting wires,reducing the effective resistance. The circuit simplifies to three resistors in series,where the middle section is shorted out. The equivalent resistance becomes $R_{after} = 1 \ \Omega + 1 \ \Omega + 1 \ \Omega = 3 \ \Omega$.The ratio of the equivalent resistances before and after the addition is $\frac{R_{before}}{R_{after}} = \frac{5 \ \Omega}{3 \ \Omega} = \frac{5}{3}$.

$A$ circuit consists of five identical conductors,each of resistance $1 \ \Omega$,as shown in the figure. Two similar conductors are added as indicated by the dotted lines. The ratio of the equivalent resistances before and after the addition is:

Similar Questions

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