Question

Four identical resistors $R$ are wired as shown in the diagram below.
What addition to this circuit described in figure will create an assembly with twice the resistance of the original circuit?

• A. Two resistors $R$ wired in parallel to each other and then connected in series to the circuit given.
• B. Two resistors $R$ wired in series to each other and then connected in parallel to the circuit given.
• C. Two resistors $R$ connected in series to the circuit given.
• D. One resistor $R$ connected in parallel to the circuit given.
• E. One resistor $R$ connected in series to the circuit given.

Answer 1

The correct option is E One resistor $R$ connected in series to the circuit given.

The equivalent resistance for the original circuit is $R_{eq}=\frac{RR}{R+R}+\frac{RR}{R+R}=R$

For A: The equivalent resistance, $R_A=\frac{RR}{R+R}+R_{eq}=R/2+R=3R/2$

For B : The equivalent resistance, $R_B=\frac{(R+R)R_{eq}}{(R+R)+R_{eq}}=\frac{2R\left(R\right)}{2R+R}=2R/3$

For C: The equivalent resistance, $R_C=(R+R)+R_{eq}=2R+R=3R$

For D: The equivalent resistance, $R_D=\frac{R{R}_{eq}}{R+R_{eq}}=\frac{RR}{R+R}=R/2$

For E: The equivalent resistance, $R_E=R+R_{eq}=R+R=2R=2R_{eq}$

So for option E we get $R_E=2R_{eq}$