
What is the equivalent resistance between points $A$ and $B$?
Answer
490.5k+ 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
Basicity of sulphurous acid and sulphuric acid are

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

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

How much time does it take to bleed after eating p class 12 biology CBSE

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

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

