Question

#Quality answer needed
Two resistors 4 Ohm and 6 Ohm are connected in parallel. The combination is connected across a 6 V battery of negligible resistance. Calculate (i) the current through the battery (ii) current through each resistor​

Answer 1

Answer:

Req = 1/R1 + 1 /R2  = 1/4 + 1/6 = 5/12 = 12/5 ohm = 2.4 ohm

According to ohm's law ,

V = IR 

I = V/R 

I = 6/2.4

I = 2.5 A

(b)  1) V = 6 V  , R= 6 ohm , I = ?

I = V/R 

I = 6/6 

I = 1 A

2) V = 6V , R = 4 ohm , I = ?

I = V/R 

I = 6 /4 

I = 1.5 A

Answer 2

Given :

Two resistors are connected in parallel combination,

R₁ = 4ohms

R₂ = 6ohms

Potential difference = 6V

To find :

The effective resistance and the current through each resistor and the current flowing through the battery.

Solution :

we know that,

The reciprocal of the combined resistance of a number of resistance connected in parallel is equal to the sum of the reciprocal of all the individual resistances. .i.e.,

1/R = 1/R₁ + 1/R₂

By substituting the values in the formula,

$\dashrightarrow \sf \dfrac{1}{R_{eq}} = \dfrac{1}{R_{1}} + \dfrac{1}{ R_2}$

$\dashrightarrow \sf \dfrac{1}{R_{eq}} = \dfrac{1}{4} + \dfrac{1}{6}$

$\dashrightarrow \sf \dfrac{1}{R_{eq}} = \dfrac{3 + 2}{12}$

$\dashrightarrow \sf \dfrac{1}{R_{eq}} = \dfrac{5}{12}$

$\dashrightarrow \sf R_{eq} = \dfrac{12}{5}$

$\dashrightarrow \sf R_{eq} = 2.4 \: ohms$

Thus, the effective resistance is 2.4 ohms.

Now, using ohms law let us find the current through the battery.

we know that,

I = VR

Where,

• I denotes current
• V denotes potential difference
• R denotes resistance

Substituting the values in the formula,

➛ I = V/R

➛ I = 6/2.4

➛ I = 2.5 A

Thus, the current through the battery is 2.5 A

Now, using ohms law let us find the current through each resistor.

Current accross R₁ :

➛ I₁ = V/R₁

➛ I₁ = 6/4

➛ I₁ = 1.5 A

Current accross R₂ :

➛ I₂ = V/R₂

➛ I₂ = 6/6

➛ I₂ = 1 A