Question

Find the effective resistance between A & B of the following figure.
A) 5 omega
B) 4 omega
C) 3 omega
D) 3 omega

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{R_{effective}=5\Omega}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given : }} \\ \tt: \implies Resistors = 2 \: \Omega \: \: (all) \\ \\ \red{\underline \bold{To \: Find : }} \\ \tt: \implies Effective \: resistance \: between \: A \: and \: B =?$

• According to given question :

$\tt \circ \: R_{1} = R_{2} =R_{3} = R_{4} = 2\Omega \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies \frac{1}{R_{parallel}} = \frac{1}{ R_{2} } + \frac{1}{ R_{3}} \\ \\ \tt: \implies \frac{1}{R_{parallel}} = \frac{1}{2} + \frac{1}{2} \\ \\ \tt: \implies \frac{1}{R_{parallel}} = \frac{1 + 1}{2} \\ \\ \tt: \implies \frac{1}{R_{parallel}} =1 \\ \\ \tt: \implies {R_{parallel}}=1 \Omega \\ \\ \bold{Again : } \\ \tt: \implies R_{effective} = R_{1} + R_{parallel} + R_{4} \\ \\ \tt: \implies R_{effective} = 2 + 1 + 2 \\ \\ \green{\tt: \implies R_{effective} = 5 \: \Omega} \\ \\ \green{\tt \therefore Effective \: resistance \: between \:A\: and \:B \: is \: 5 \: \Omega}$

Answer 2

Looking at the attachment we can say that, R1, R2, R3 and R4 are 2 ohm.

R1 = R2 = R3 = R4 = 2 ohm

Both R2 and R3 are connected in parallel. And for parallel combination, formula is 1/Rp = 1/R2 + 1/R3

→ 1/Rp = 1/2 + 1/2

→ 1/Rp = 2/2

→ Rp = 1 ohm

Now, we left with R1, Rp and R4. And all the three are connected in series.

For series combination,

→ Rs = R1 + Rp + R4

→ Rp = 2 + 1 + 2

→ Rp = 5 ohm

As per given condition we have to find the effective resistance between A & B.

And from above calculations effective resistance is equal to Rp.

Hence, the effective resistance between A and B is 5 ohm/omega.

Option a) 5 omega