IMAT 2016 Q57 [Resistance]

A student has three 6.0 Ω resistors that can be connected together in any configuration. What are the maximum and minimum resistances that can be obtained by using one or more of these three resistors?

[Assume the connections between the resistors have negligible resistance, the temperature of the resistors is constant, the resistors are used in a d.c. circuit and none of the resistors is short-­circuited.]

A. maximum resistance: 6.0 Ω; minimum resistance: 2.0 Ω
B. maximum resistance: 12 Ω; minimum resistance: 0.50 Ω
C. maximum resistance: 6.0 Ω; minimum resistance: 0.50 Ω
D. maximum resistance: 18 Ω; minimum resistance: 6.0 Ω
E. maximum resistance: 18.0 Ω; minimum resistance: 2.0 Ω

The given resistors will be used as R=6\Omega

Certain rules need to be remembered:

a) n resistors connected in series: R_e = R_1+R_2+\dotsc + R_n

b)m resistors connected in parallel: \frac{1}{R_e} =\frac{1}{R_1}+\frac{1}{R_2} + \dotsc +\frac{1}{R_m}

By looking at the rules, we can figure out that the maximum resistance can be obtained by connecting resistors in series, whereas the minimum can be obtained when connecting them in parallel.

  1. maximum resistance (connected in series)

R_{max}=R +R+R = 3R = 3\cdot6\Omega =18\Omega

  1. minimum resistance (connected in parallel)

\frac{1}{R_{min}} = \frac{1}{R} +\frac{1}{R}+\frac{1}{R}=\frac{3}{R} \to R_{min} = \frac{R}{3} = \frac{6\Omega}{3} = 2\Omega

Concluding that (E) is the correct answer.