Question

Three resistors $$(2\mathrm{\Omega },4\mathrm{\Omega },4\mathrm{\Omega })$$ are combined to achieve $$R_{max}$$ (maximum resistance) and $$R_{min}$$ (minimum resistance). Then the average and ratio of $$R_{max}$$ and $$R_{min}$$ shall be respectively:

A. 5.5, 1:10
B. 5.5, 11:1
$$C.{\displaystyle \frac{22}{4}},10:1$$
D. Both 1 and 3 are correct

Answer 1

Hint: We will make use of the formula for calculating the resistors connected in parallel to find the value of the minimum resistance and the formula for calculating the resistors connected in series to find the value of the maximum resistance. Then, we will find the average and the ratio of these values.
Formula used:
$${\displaystyle \frac{1}{R_P}}={\displaystyle \frac{1}{R_1}}+{\displaystyle \frac{1}{R_2}}+{\displaystyle \frac{1}{R_3}}+......+{\displaystyle \frac{1}{R_n}}$$
$$R_S=R_1+R_2+R_3+......+R_n$$

Complete answer:
The formula used to find the resistance of the resistors connected in parallel is as follows.
$${\displaystyle \frac{1}{R_P}}={\displaystyle \frac{1}{R_1}}+{\displaystyle \frac{1}{R_2}}+{\displaystyle \frac{1}{R_3}}+......+{\displaystyle \frac{1}{R_n}}$$
Similarly, the formula used to find the resistance of the resistors connected in series is as follows.
$$R_S=R_1+R_2+R_3+......+R_n$$
From given, we have the values of the resistors to be equal to$$(2\mathrm{\Omega },4\mathrm{\Omega },4\mathrm{\Omega })$$.
As we are given with the resistance values of the 3 resistors, firstly, we will derive the formula for the equivalent resistance of the 3 resistors connected in parallel.
So, we have,
$${\displaystyle \frac{1}{R_P}}={\displaystyle \frac{1}{R_1}}+{\displaystyle \frac{1}{R_2}}+{\displaystyle \frac{1}{R_3}}$$
Now, substitute the values of the resistance of the 3 resistors given in the above equation.
$$\begin{array}{rl}& {\displaystyle \frac{1}{R_P}}={\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{4}}\\ & \Rightarrow {\displaystyle \frac{1}{R_P}}={\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2}}\end{array}$$
Continue the further calculation.
$$\begin{array}{rl}& {\displaystyle \frac{1}{R_P}}=1\\ & \Rightarrow R_P=1\end{array}$$
Therefore, the value of the minimum resistance is$$1\mathrm{\Omega }$$.
As we are given with the resistance values of the 3 resistors, firstly, we will derive the formula for the equivalent resistance of the 3 resistors connected in series.
So, we have,
$$R_S=R_1+R_2+R_3$$
Now, substitute the values of the resistance of the 3 resistors given in the above equation.
$$\begin{array}{rl}& R_S=2+4+4\\ & \Rightarrow R_S=10\end{array}$$
Therefore, the value of the maximum resistance is$$10\mathrm{\Omega }$$.
The average value of the minimum resistance $$1\mathrm{\Omega }$$ and the maximum resistance $$10\mathrm{\Omega }$$ is calculated as follows.
$$\begin{array}{rl}& \text{Avg}={\displaystyle \frac{R_{max}+R_{min}}{2}}\\ & \Rightarrow \text{Avg}={\displaystyle \frac{10+1}{2}}\end{array}$$
Upon further continuing the calculation, we get,
 $$\text{Avg}=5.5$$
Therefore, the average value of the maximum and minimum resistance is 5.5 or $${\displaystyle \frac{22}{4}}$$.
The ratio of the minimum resistance $$1\mathrm{\Omega }$$and the maximum resistance $$10\mathrm{\Omega }$$ is calculated as follows.
$$\begin{array}{rl}& R={\displaystyle \frac{R_{max}}{R_{min}}}\\ & \Rightarrow R={\displaystyle \frac{10}{1}}\end{array}$$
Therefore, the ratio of the maximum and minimum resistance is 10:1.
As the values the average and ratio of $$R_{max}$$ and $$R_{min}$$ are $${\displaystyle \frac{22}{4}},10:1$$.

So, the correct answer is “Option C”.

Note:
The average of the values of the maximum and minimum resistance can be found by first adding the maximum and minimum values and then dividing this value by 2. The ratio of the values of the maximum and minimum resistance can be found by dividing the values of the maximum and minimum resistance.