Question

consider the circuit shown in figure. Calculate the current through the 3 ohm resistor ​

Answer 1

Answer:

the ans is 1.33 Amp of your question

Answer 2

Answer:

The current through the 3-ohm resistor ​will be 1.33A.

Explanation:

There are three resistors given in the question i.e.

R₁=4Ω

R₂=3Ω

R₃=6Ω

Of which R₂ and R₃ connected in parallel and this parallel circuit is connected in series with R₁. So, first, let's find the equivalent resistance of the parallel circuit.

$\frac{1}{R_p}=\frac{1}{R_2}+\frac{1}{R_3}$             (1)

By substituting the values of R₂ and R₃ in the equation (1) we get;

$\frac{1}{R_p}=\frac{1}{3}+\frac{1}{6}$

$R_p=\frac{6}{3}=2\Omega$              (2)

The equivalent resistance of the whole circuit will be,

$R_e_q=R_1+R_p$               (3)

By substituting the value of R₁ and equation (2) in equation (3) we get;

$R_e_q=4+2=6\Omega$  (4)

The current in the circuit is given as,

$I=\frac{V}{R_e_q}$         (5)

By substituting the value of V=12 v and equation (4) in equation (5) we get;

$I=\frac{12}{6}=2A$               (6)

The voltage across the 4Ω resistor is given as,

$V_4=I\times R_1$

So,

$V_4=2\times 4=8V$          (7)

The voltage left for the parallel circuit is;

$V_p=V-V_4$

$V_p=12-8=4V$      (8)

As the resistors, R₂ and R₃ are connected in parallel the voltage(Vp) across them remains constant i.e. 4V

Now the current(I₃) through 3Ω resistor is given as;

$I_3=\frac{V_p}{R_2}$                (9)

By substituting the values of  Vp and R₂ in equation (9) we get;

$I_3=\frac{4}{3}=1.33 A$

Hence, the current through the 3-ohm resistor ​will be 1.33A.