Question

Question number 30 pls

Answer 1

Answer:

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Answer 2

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The current passing through the 4 ohm resistor in the circuit is \dfrac{3}{4}.

Explanation:

Given that,

Resistance R_{1}= 2 \Omega

Resistance R_{2}= 4 \Omega

Resistance R_{3}= 2 \Omega

Resistance R_{4}= 6 \Omega

Voltage V = 5 V

R₁ and R₂ is connected in series,

The resistance is

R'= R_{1}+R_{2}

R'=2+4

R'=6\ ohm

Now, R' and R₄ is connected in parallel,

\dfrac{1}{R''}=\dfrac{1}{R'}+\dfrac{1}{R_{4}}

\dfrac{1}{R''}=\dfrac{1}{6}+\dfrac{1}{6}

R''=3\ ohm

Now, R'' and R₃ is connected in series

R'''=R''+R_{3}

R'''=3+2

R'''=5\ ohm

Using Ohm's law

V= IR

The net current is defined as:

I=\dfrac{V}{R}

Here, V = voltage

R = equivalent resistance

I = \dfrac{5}{5}

I = 1\ A

The current passing through the 4 ohm resistor in the circuit

I =\dfrac{I_{net}\times R''}{R_{2}}

I = \dfrac{1\times 3}{4}

I = \dfrac{3}{4}

Hence, The current passing through the 4 ohm resistor in the circuit is \dfrac{3}{4}.