
Determine the equivalent resistance of the following network.
Answer
564.3k+ views
Hint
The above network of resistors is a series combination of single identical units. The same unit is connected in series four times. To solve the problem, we find the equivalent resistance of the single unit and multiply it by four, because four such units are connected in series. To find the equivalent resistance of a single unit, we use the formula for finding equivalent resistance in parallel as well as series.
Equivalent resistance in parallel, ${R_{p - equ}} = \dfrac{1}{{\dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}... + \dfrac{1}{{{R_n}}}}}$
Equivalent resistance in series, ${R_{s - equ}} = {R_1} + {R_2} + ... + {R_n}$
Here, equivalent resistance in parallel is represented by ${R_{p - equ}}$
Equivalent resistance in series is represented by ${R_{s - equ}}$
Different resistors are represented by ${R_1},{R_2},...{R_n}$
Complete step by step answer
The network of resistors in the question is nothing but a repetition of a single unit in series.
The given network can also be drawn as
The resistance on the top as well as the bottom of a single unit is in series, so you directly add the resistors to get equivalent resistance of the two resistors at the top and bottom.
$\Rightarrow {R_{equ - top}} = 1 + 1 = 2 $
$\Rightarrow {R_{equ - bottom}} = 2 + 2 = 4$
Then use the formula for finding the equivalent resistance in parallel. This way we find the equivalent resistance for a single unit.
$\Rightarrow {R_{p - equ}} = \dfrac{1}{{\dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}}} = \dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} $
$\Rightarrow {\operatorname{R} _{p - equ}} = \dfrac{{2 \times 4}}{{2 + 4}} = \dfrac{4}{3} $
The network has four such units connected in series, so we simply add the same equivalent resistance four times or multiply it by four to get the equivalent resistance of the entire network of resistors.
$\Rightarrow {R_{equ}} = 4 \times \dfrac{4}{3} = \dfrac{{16}}{3}\Omega $
The total equivalent resistance is${R_{equ}} = \dfrac{{16}}{3}\Omega $
Units of resistance are ohms.
Note
Resistors are said to be connected in parallel if their terminals are connected to the same two nodes. The equivalent overall resistance is smaller than the smallest parallel resistor.
Resistors are said to be connected in series when they are connected head-to-tail and no other wires are branching off from the nodes between components. Their equivalent resistance is simply the sum of all the resistors connected in series.
The above network of resistors is a series combination of single identical units. The same unit is connected in series four times. To solve the problem, we find the equivalent resistance of the single unit and multiply it by four, because four such units are connected in series. To find the equivalent resistance of a single unit, we use the formula for finding equivalent resistance in parallel as well as series.
Equivalent resistance in parallel, ${R_{p - equ}} = \dfrac{1}{{\dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}... + \dfrac{1}{{{R_n}}}}}$
Equivalent resistance in series, ${R_{s - equ}} = {R_1} + {R_2} + ... + {R_n}$
Here, equivalent resistance in parallel is represented by ${R_{p - equ}}$
Equivalent resistance in series is represented by ${R_{s - equ}}$
Different resistors are represented by ${R_1},{R_2},...{R_n}$
Complete step by step answer
The network of resistors in the question is nothing but a repetition of a single unit in series.
The given network can also be drawn as
The resistance on the top as well as the bottom of a single unit is in series, so you directly add the resistors to get equivalent resistance of the two resistors at the top and bottom.
$\Rightarrow {R_{equ - top}} = 1 + 1 = 2 $
$\Rightarrow {R_{equ - bottom}} = 2 + 2 = 4$
Then use the formula for finding the equivalent resistance in parallel. This way we find the equivalent resistance for a single unit.
$\Rightarrow {R_{p - equ}} = \dfrac{1}{{\dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}}} = \dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} $
$\Rightarrow {\operatorname{R} _{p - equ}} = \dfrac{{2 \times 4}}{{2 + 4}} = \dfrac{4}{3} $
The network has four such units connected in series, so we simply add the same equivalent resistance four times or multiply it by four to get the equivalent resistance of the entire network of resistors.
$\Rightarrow {R_{equ}} = 4 \times \dfrac{4}{3} = \dfrac{{16}}{3}\Omega $
The total equivalent resistance is${R_{equ}} = \dfrac{{16}}{3}\Omega $
Units of resistance are ohms.
Note
Resistors are said to be connected in parallel if their terminals are connected to the same two nodes. The equivalent overall resistance is smaller than the smallest parallel resistor.
Resistors are said to be connected in series when they are connected head-to-tail and no other wires are branching off from the nodes between components. Their equivalent resistance is simply the sum of all the resistors connected in series.
Recently Updated Pages
A man running at a speed 5 ms is viewed in the side class 12 physics CBSE

The number of solutions in x in 02pi for which sqrt class 12 maths CBSE

State and explain Hardy Weinbergs Principle class 12 biology CBSE

Write any two methods of preparation of phenol Give class 12 chemistry CBSE

Which of the following statements is wrong a Amnion class 12 biology CBSE

Differentiate between action potential and resting class 12 biology CBSE

Trending doubts
What are the major means of transport Explain each class 12 social science CBSE

Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

Explain sex determination in humans with line diag class 12 biology CBSE

Explain sex determination in humans with the help of class 12 biology CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

