Question

In the circuit shown
here, calculate the
equivalent resistance
between the points A
and B.
​

Answer 1

Given :

▪ A circuit containing four idential resistors has been provided.

To Find :

▪ Eq. resistance b/w A and B.

Formula :

✴ Eq. resistance of series :

$\bigstar\:\underline{\boxed{\bf{\red{R_s=R_1+R_2+...+R_n}}}}$

✴ Eq. resistance of parallel :

$\bigstar\:\underline{\boxed{\bf{\blue{\dfrac{1}{R_p}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+...+\dfrac{1}{R_n}}}}}$

Calculation :

☣ Eq. resistance of lower part :

All three resistors (2Ω, 3Ω, and 4Ω) are connected in series.

------|R_1|-----|R_2|-----|R_3|-----

$\dashrightarrow\sf\:R=R_1+R_2+R_3\\ \\ \dashrightarrow\sf\:R=2+3+4\\ \\ \dashrightarrow\bf\:R=9\Omega$

☣ Net eq. resistance of circuit :

R and R_4 are connected in parallel.

---------------|R_4|-------------

---------------| R |----------------

$\longrightarrow\sf\:\dfrac{1}{R_{eq}}=\dfrac{1}{R_4}+\dfrac{1}{R}\\ \\ \longrightarrow\sf\:\dfrac{1}{R_{eq}}=\dfrac{1}{9}+\dfrac{1}{9}=\dfrac{2}{9}\\ \\ \longrightarrow\sf\:R_{eq}=\dfrac{9}{2}\\ \\ \longrightarrow\underline{\boxed{\bf{\gray{R_{eq}=4.5\Omega}}}}\:\orange{\bigstar}$