
What is the equivalent resistance between points $A$ and $B$?
Answer
505.8k+ views
Hint: The cumulative resistance of a series circuit is precisely the sum of the resistances of the circuit's components. Since current will pass through several pathways in a parallel circuit, the total overall resistance is lower than the resistance of any single part.
Formula used:
The formula to calculate resistance in series combination is:
$R = {R_1} + {R_2} + {R_3} + .... + {R_n}$
And the formula to calculate the resistance in parallel combination is:
$\dfrac{1}{R} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} + ... + \dfrac{1}{{{R_n}}}$
Complete step by step answer:
To calculate the total resistance , we will divide the circuits into different parts. Firstly, we combine ${R_3}$ and ${R_4}$, these combinations are in series.
Hence, we sum up the values.
$70 + 30 = 100\Omega \\ $
Now, this $100\Omega $ is in parallel combination with ${R_2}$.Hence,
$\dfrac{1}{R} = \dfrac{1}{{100}} + \dfrac{1}{{100}} \\
\Rightarrow \dfrac{1}{R} = \dfrac{{1 + 1}}{{100}} \\
\Rightarrow \dfrac{1}{R} = \dfrac{2}{{100}} \\
\Rightarrow R = 50\Omega \\ $
Now, The resistance $50\Omega $ is in parallel with ${R_5}$. Hence, we will get,
$\dfrac{1}{R} = \dfrac{1}{{50}} + \dfrac{1}{{50}} \\
\Rightarrow \dfrac{1}{R} = \dfrac{2}{{50}} \\
\Rightarrow R = 25\Omega \\ $
Now, this $25\Omega $ resistance is in series with ${R_1}$.
Hence we get,
$R = 50 + 25 \\
\therefore R = 75\Omega \\ $
Hence, the equivalent resistance between point A and B is $75\Omega $.
Note: Resistor Combination or mixed resistor circuits are resistor circuits that incorporate series and parallel resistor networks together. The procedure for measuring the equivalent resistance of the circuit is the same as for any other series or parallel circuit.
Formula used:
The formula to calculate resistance in series combination is:
$R = {R_1} + {R_2} + {R_3} + .... + {R_n}$
And the formula to calculate the resistance in parallel combination is:
$\dfrac{1}{R} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} + ... + \dfrac{1}{{{R_n}}}$
Complete step by step answer:
To calculate the total resistance , we will divide the circuits into different parts. Firstly, we combine ${R_3}$ and ${R_4}$, these combinations are in series.
Hence, we sum up the values.
$70 + 30 = 100\Omega \\ $
Now, this $100\Omega $ is in parallel combination with ${R_2}$.Hence,
$\dfrac{1}{R} = \dfrac{1}{{100}} + \dfrac{1}{{100}} \\
\Rightarrow \dfrac{1}{R} = \dfrac{{1 + 1}}{{100}} \\
\Rightarrow \dfrac{1}{R} = \dfrac{2}{{100}} \\
\Rightarrow R = 50\Omega \\ $
Now, The resistance $50\Omega $ is in parallel with ${R_5}$. Hence, we will get,
$\dfrac{1}{R} = \dfrac{1}{{50}} + \dfrac{1}{{50}} \\
\Rightarrow \dfrac{1}{R} = \dfrac{2}{{50}} \\
\Rightarrow R = 25\Omega \\ $
Now, this $25\Omega $ resistance is in series with ${R_1}$.
Hence we get,
$R = 50 + 25 \\
\therefore R = 75\Omega \\ $
Hence, the equivalent resistance between point A and B is $75\Omega $.
Note: Resistor Combination or mixed resistor circuits are resistor circuits that incorporate series and parallel resistor networks together. The procedure for measuring the equivalent resistance of the circuit is the same as for any other series or parallel circuit.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Class 12 Question and Answer - Your Ultimate Solutions Guide

Complete reduction of benzene diazonium chloride with class 12 chemistry CBSE

How can you identify optical isomers class 12 chemistry CBSE

Trending doubts
Which are the Top 10 Largest Countries of the World?

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

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

Differentiate between insitu conservation and exsitu class 12 biology CBSE

Draw a neat and well labeled diagram of TS of ovary class 12 biology CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

