Respuesta :
Answer:
In series, total resistance = 1000001Ω
In parallel, the total resistance = 1.0 Ω
Explanation:Note that:
• When two, resistors,, R₁ and R₂ are connected in ,series,, the total resistance is the, sum, of their individual resistances
That is, Total resitance = R₁ + R₂
• When two, resistors,, R₁ and R₂ are connected in ,parallel,, the inverse of the total resistance equals the ,sum of the inverse, of the individual resistances.
[tex]\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}[/tex]For 1.0 Ω and 1.0 x 10⁶Ω resistors:
R₁ = 1.0 Ω
R₂ = 1.0 x 10⁶Ω
If the resistors are connected in series
[tex]\begin{gathered} R_T=R_1+R_2 \\ R_T=1.0\Omega\text{ + 1.0}\times10^6\Omega \\ R_T=1000001\Omega \\ \text{Total resistance = }1000001\Omega \end{gathered}[/tex]The 1 x 10⁶Ω resistor dominates because it is largely greater than the 1.0 Ω resistor. Since the two resistors are connected in series, the 1 x 10⁶Ω resistor will dominate
If the resistors are connected in parallel
[tex]\begin{gathered} \frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2} \\ \frac{1}{R_T}=\frac{1}{1.0}+\frac{1}{1\times10^6} \\ \frac{1}{R_T}=1+10^{-6} \\ \frac{1}{R_T}=1.000001 \\ R_T=\frac{1}{1.000001} \\ R_T=1.0\Omega \\ \text{Total resistance =1.0}\Omega \end{gathered}[/tex]The 1.0 Ω resistor dominates because it is way smaller than the 1 x 10⁶Ω resistor. Because the resistors are connected in parallel, and the inverse of each of the resistors, are taken, the 1.0 Ω resistor has to dominate.
