Question

calculate the equivalent resistance in between the point a and b of the circuit shown in figure a and b (final answer a.2.4 ohm b.4.8 ohm)​

Answer 1

Solution :-

First of all we should know about the Parallel and Series circuit.

Series circuit :- When the current across the Resistance is same then they are in Series circuit.

Parallel circuit :- When the potential difference across the Resistance is same then they are in Parallel circuit.

**Refer the attachment for the reduced circuit.

For (a)

After reducing the circuit we are able to see 3Ω And 12Ω in parallel circuit.

Now as in parallel circuit :-

$R_{eq} = \left( \dfrac{1}{r_1} + \dfrac{1}{r_2} + \dfrac{1}{r_3} .... + \dfrac{1}{r_n}\right)^{-1}$

So our equivalent Resistance

$R_{eq} = \left( \dfrac{1}{3} + \dfrac{1}{12} \right)^{-1}$

$= \left( \dfrac{4 + 1}{12} \right)^{-1}$

$= \left( \dfrac{5}{12}\right)^{-1}$

$= \dfrac{12}{5}$

$= 2.4 \Omega$

For (b)

After reducing the circuit we are able to see 6Ω And 24Ω in parallel circuit.

Now as in parallel circuit :-

$R_{eq} = \left( \dfrac{1}{r_1} + \dfrac{1}{r_2} + \dfrac{1}{r_3} .... + \dfrac{1}{r_n}\right)^{-1}$

So our equivalent Resistance

$R_{eq} = \left( \dfrac{1}{6} + \dfrac{1}{24} \right)^{-1}$

$= \left( \dfrac{4 + 1}{24} \right)^{-1}$

$= \left( \dfrac{5}{24}\right)^{-1}$

$= \dfrac{24}{5}$

$= 4.8 \Omega$