Question

Find the current passing through the 4 ohm resistor in the given circuit

Answer 1

according to ohm.s law v=IR
here l=v/r
and hence l= 5/6 ampere

Answer 2

Answer:

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 $\dfrac{3}{4}$.