Question

A 1.0 Ω resistor is hooked up to a 1.0 x106 Ω resistor in series and then in parallel. Calculate the total resistance for each situation and explain the dominance of one resistor.

Answer 1

Answer:

In series, total resistance = 1000001Ω

In parallel, the total resistance = 1.0 Ω

Step-by-step 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.

`(1)/(R_T)=(1)/(R_1)+(1)/(R_2)`

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

`\begin{gathered} R_T=R_1+R_2 \\ R_T=1.0\Omega\text{ + 1.0}*10^6\Omega \\ R_T=1000001\Omega \\ \text{Total resistance = }1000001\Omega \end{gathered}`

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

`\begin{gathered} (1)/(R_T)=(1)/(R_1)+(1)/(R_2) \\ (1)/(R_T)=(1)/(1.0)+(1)/(1*10^6) \\ (1)/(R_T)=1+10^(-6) \\ (1)/(R_T)=1.000001 \\ R_T=(1)/(1.000001) \\ R_T=1.0\Omega \\ \text{Total resistance =1.0}\Omega \end{gathered}`

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.