Question

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Answer 1

Answer:

$\frac{33}{4}$Ω

Explanation:

Given:

Resistors

R₁=R₂=R₃=R₄=R₅=R₆= 3ohms

To find:

Net resistance across the circuit

We know that resistors R₂,R₃ and R₄ are in series,

When any two resistors are in series, equivalent resistance = R₁+R₂

R₂+R₃+R₄ = 3+3+3 = 9 ohms (let this be called as R')

Now,

R' becomes parallel to R₅

When any two resistors are in parallel, equivalent resistance = $\frac{1}{R1} +\frac{1}{R2}$

So here, equivalent resistance = $\frac{1}{R}=\frac{1}{9} +\frac{1}{3}$

$\frac{1}{R}=\frac{1}{9} +\frac{3}{9}$

$\frac{1}{R}=\frac{4}{9}$

R=$\frac{9}{4}$ (let this be called as R*)

R* is in series with R₁ and R₆, so:

R (eq) = $\frac{9}{4}+3+3$

R (eq) = $\frac{9}{4}+6$

R (eq) = $\frac{9}{4} +\frac{24}{4}$

R (eq) = $\frac{33}{4} \ ohms$

The net resistance across the circuit is equal to $\frac{33}{4}$Ω

Answer 2

Answer:

33/4∩

Explanation:

hope it is helpful