Question

What is the equivalent resistance between points $A$ and $B$?

Answer 1

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.