Question

Three identical resistors have an equivalent resistance of 85 ω when connected in parallel. part a what is their equivalent resistance when connected in series?

Answer 1

The equivalent resistance of the three resistors when connected in parallel is:
$\frac{1}{R_{eq}} = \frac{1}{R_1}+ \frac{1}{R_2}+ \frac{1}{R_3}$
Since the three resistors in this problem are identical, we can call R their resistance, and we can rewrite the previous equation as
$\frac{1}{R_{eq}} = \frac{1}{R}+ \frac{1}{R}+ \frac{1}{R}= \frac{3}{R}$
And since we know the value of the equivalent resistance, $R_{eq}=85 \Omega$, we can find the value of R:
$R=3 R_{eq} = 3 \cdot 85 \Omega=255 \Omega$

Now the problem asks us what is the equivalent resistance of the three resistors when they are connected in series. In this case, the equivalent resistance is just the sum of the three resistances, so
$R_{eq} = R+R+R=3 R= 3 \cdot 255 \Omega = 765 \Omega$

Answer 2

Answer:

90 for when Ω = 10 (Another Variation for Problem)

Explanation:

3 * 3 * 10