Question

pls. solve urgently....

no scam pls....

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Answer 1

(a) Effective resistance of two 8 ohm resistors in the combination

As it is a parallel combination ,we'll use :

$\boxed{\green{\frac{1}{Rp}=\frac{1}{R1}+\frac{1}{R2}}}$

$\displaystyle{\implies \frac{1}{Rp}=\frac{1}{8}+\frac{1}{8}}$

$\displaystyle{\implies \frac{1}{Rp}=\frac{2}{8}}$

$\displaystyle{\implies \frac{1}{Rp}= \frac{\cancel 2}{\cancel 8} \implies \frac{1}{Rp}=\frac{1}{4}}$

$\bf{\orange{\implies Rp= 4\; \Omega}}$

(b) Current flowing through 4 ohm

resistor

$\boxed{\pink{V=IR}}$

where,

V refers to potential difference

I refers to electic current

R refers to resistance

$\displaystyle{\implies 8=I \times 4}$

$\displaystyle{\implies \frac{8}{4}=I \implies I=\frac{8}{4}}$

$\bf{\orange{\implies I = 2\;ampere}}$

Answer 2

Answer:

answer will be the 4ohm and 2amp

Explanation:

here 2 resistors of 8ohm are connected in parralel to each other so effective resistance will be :

1/Rs= 1/R1+1/R2

so, 1/Rs= 1/8 +1/8

1/Rs= 1+1/8

1/Rs= 2/8

1/Rs=1/4

Rs= 4ohm

so the effective resistance of both the resistors is 8 ohm

now we have to find the current flowing through resistor of 4 ohm,

from ohm's law we know that,

V=I×R

so applying the formula we get,

8=I×4

I=8/4

I=2amp

so current flowing through resistir if 4 ohm from a battery of 8v is 2 ampere.

HOPE IT WILL BE HELPFUL !!