Question

problem 19.21
two resistors when connected in series to a 120 - v line use one - fourth the power that is used when they are connected in parallel.
part a
if one resistor is 4.2 kω, what is the resistance of the other?
express your answer to two significant figures and include the appropriate units.
r₂ = value units

Explanation:

Step1: Recall power - resistance formula

The power formula is $P=\frac{V^{2}}{R}$. Let the two resistors be $R_1$ and $R_2$. The equivalent resistance in series $R_s=R_1 + R_2$, and in parallel $R_p=\frac{R_1R_2}{R_1 + R_2}$. Given $P_s=\frac{1}{4}P_p$. Since $P=\frac{V^{2}}{R}$, we have $\frac{V^{2}}{R_s}=\frac{1}{4}\times\frac{V^{2}}{R_p}$.

Step2: Simplify the power - resistance relationship

Canceling out $V^{2}$ from both sides of $\frac{V^{2}}{R_s}=\frac{1}{4}\times\frac{V^{2}}{R_p}$, we get $\frac{1}{R_1 + R_2}=\frac{1}{4}\times\frac{R_1 + R_2}{R_1R_2}$. Cross - multiplying gives $4R_1R_2=(R_1 + R_2)^2$. Expanding the right - hand side: $4R_1R_2=R_1^{2}+2R_1R_2+R_2^{2}$. Rearranging, we have $R_1^{2}-2R_1R_2 + R_2^{2}=0$, which is $(R_1 - R_2)^2 = 0$, so $R_1=R_2$.

Step3: Calculate the value of $R_2$

Given $R_1 = 4.2\ k\Omega$, since $R_1 = R_2$, then $R_2=4.2\ k\Omega$.

Answer:

$4.2\ k\Omega$