Three resistors, 4.0-$\varOmega$, 8.0-$\varOmega$, 16-$\varOmega$, are connected in parallel in a circuit. What is the equivalent resistance of this combination of resistors?
A)30 $\varOmega$
B)10 $\varOmega$
C)2.3 $\varOmega$
D)2.9$\varOmega$
E)0.34 $\varOmega$

Question

Quizplus · Accepted Answer

The equivalent resistance $ R_{eq} $ for resistors in parallel is given by the formula: $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$. Substituting the given values: $\frac{1}{R_{eq}} = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} = \frac{4+2+1}{16} = \frac{7}{16}$. Therefore, $ R_{eq} = \frac{16}{7} \approx 2.3 \, \varOmega $.

Three Resistors, 4 Ω\varOmegaΩ , 80- Ω\varOmegaΩ , 16- Ω\varOmegaΩ , Are Connected in Parallel in a Circuit

Three Resistors, 4 $\varOmega$ , 80- $\varOmega$ , 16- $\varOmega$ , Are Connected in Parallel in a Circuit