Question

the current I in the circuit shown below is-

Answer 1

R equivalent in figure is in series with 4 ohm resistor placed adjacent to 6 V battery.Therefore, effective resistance =6 ohm.

Answer 2

Answer:

The current I in the circuit is 0.5 A.

(3) is correct.

Explanation:

Given that,

$R_{1}=4\Omega$

$R_{2}=4\Omega$

$R_{3}=4\Omega$

$R_{4}=2\Omega$

$R_{5}=4\Omega$

According to figure,

Firstly , R₂ and R₃ is connected in parallel

The equivalent resistance is

$\dfrac{1}{R'}=\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}$

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

$\dfrac{1}{R'}=\dfrac{2}{4}$

$R'=2\Omega$

Now, R' and R₄ is connected in series

The equivalent resistance is

$R''=R'+R_{4}$

$R''=2+2$

$R''=4\Omega$

Now, R'' and R₅ is connected in parallel

The equivalent resistance is

$\dfrac{1}{R'''}=\dfrac{1}{R''}+\dfrac{1}{R_{5}}$

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

$\dfrac{1}{R'''}=\dfrac{2}{4}$

$R'''=2\Omega$

Now, R₁ and R''' is connected in series

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

$R = 4+2$

$R = 6\Omega$

Using Ohm's law

The current in the circuit is

$V = IR$

$I =\dfrac{V}{I}$

$I = \dfrac{6}{6}$

$I = 1\ A$

We need to the calculate the potential through first resistor

$V' =\dfrac{I}{R}$

$V' = 1\times4$

$V' = 4\ V$

Now, The potential difference is

$V''=V-V'$

$V'' = 6V-4V$

$V''=2 V$

Therefore,

The potential difference will be 2v in middle circuit.

Therefore, The current in the circuit will be

$I =\dfrac{V}{R}$

$I=\dfrac{2}{4}$

$I = 0.5\ A$

Hence, The current I in the circuit is 0.5 A.