Question

suppose that you have a 640 - ω, a 790 - ω, and a 1.50 - kω resistor. part a what is the maximum resistance you can obtain by combining these? express your answer to three significant figures and include the appropriate units. part b what is the minimum resistance you can obtain by combining these? express your answer to three significant figures and include the appropriate units.

Explanation:

Step1: Recall resistance - combination rules

The maximum resistance is obtained when resistors are connected in series. The formula for equivalent resistance in series is $R_{eq}=R_1 + R_2+R_3$. First, convert $1.50\ k\Omega$ to $\Omega$. Since $1\ k\Omega = 1000\ \Omega$, then $1.50\ k\Omega=1500\ \Omega$.

Step2: Calculate maximum resistance

$R_{max}=640\ \Omega + 790\ \Omega+1500\ \Omega=2930\ \Omega$

Step3: Recall minimum - resistance rule

The minimum resistance is obtained when resistors are connected in parallel. The formula for equivalent resistance in parallel is $\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$.

Step4: Calculate minimum resistance

$\frac{1}{R_{min}}=\frac{1}{640\ \Omega}+\frac{1}{790\ \Omega}+\frac{1}{1500\ \Omega}$
$\frac{1}{R_{min}}=\frac{790\times1500 + 640\times1500+640\times790}{640\times790\times1500}$
$\frac{1}{R_{min}}=\frac{1185000+960000 + 505600}{640\times790\times1500}=\frac{2650600}{758400000}$
$R_{min}=\frac{758400000}{2650600}\approx286\ \Omega$

Answer:

Part A:
Value: $2930$
Units: $\Omega$
Part B:
Value: $286$
Units: $\Omega$