Question

In the circuit shown in figure the current flowing through 5 Ω resistance is :​

Answer 1

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Req= 10×25/(10+25)

=50/7

Using V=iReq

V=2.1×50/7= 15 volt

Now at 5 ohm

V=iR

i=v/R=15/25=0.6

Current at 5 ohm resistance is 0.6 ampere

Answer 2

Concept:

When the resistances are in series, the resultant can be calculated by the algebraic sum of the individual resistances.

Given:

A circuit is given.

Find:

The current flows through 5Ω resistance.

Solution:

In above circuit,

R = R₁ +R₂

R = 8 + 2 = 10Ω

Similarly, in lower circuit,

R' = R₁ +R₂

R' = 20 + 5 = 25Ω

In a parallel combination of resistor, we know that current through R' can be calculated by,

I₁ = [(R)/(R+R')] I

I₁ = [(10)/(10+25)] × 2.1

I₁ = 10/35 × 2.1

I₁ = 0.6 A

As the 20Ω  and 5Ω are in series, the current flowing through will be the same.

Hence, the current flowing through 5 Ω resistance is 0.6A.

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