Question

three resistors of 3 ohm each are connected to a battery 3V as Shown. calculate the current drawn from the battery.

Answer 1

Given values,

R1=3ohms

R2=3ohms

R3=3ohms

V=3volt

R4=R1+R2

R4=3ohms+3ohms

R4=6ohms

1/R5=1/R4+1/R3

1/R5=1/6+1/3

1/R5=1+2/6

1/R5=3/6

1/R5=1/2

R5=2ohms

Current (I)=V/R5

I. =3volt/2ohms

I. =1.5 amper(A)

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Answer 2

Answer:

Current, I = 1.5 A

Explanation:

Resistance, $R_1=R_2=R_3=3\ \Omega$

Voltage of the battery, V = 3 V

Firstly calculating the equivalent resistance of the circuit. The two resistors are in series.

R' = 6 ohms

Now R' and R₃ are in parallel. Its equivalent is given by :

$\dfrac{1}{R_{eq}}=\dfrac{1}{6}+\dfrac{1}{3}$

$R_{eq}=2\ \Omega$

Let I is the current flowing for the battery. Using the Ohm's law as :

V = I R

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

$I=\dfrac{3}{2}$

I = 1.5 A

So, the current drawn from the battery is 1.5 A. Hence, this is the required solution.