Question

Four resistors of 12, 3.0, 5.0, and 4.0 Ω are connected in parallel. A 12-V battery is connected to the combination. What is the current through the battery?

Answer 1

Therefore,

The current through the battery is 10.4 Ampere.

Step-by-step explanation:

Given:

V = 12 V Battery

Connection is Parallel,

R₁ = 12 Ω

R₂ = 3 Ω

R₃ = 5 Ω

R₄ = 4 Ω

Let I₁, I₂, I₃, I₄ be the current passing through R₁ ,R₂, R₃, R₄ resistors

To Find:

I =? (current through the battery)

Solution:

As Connection is Parallel Voltage Remain SAME through resistors

Bu Ohm's Law we have

`I =(V)/(R)`

So current through R₁

`I_(1)=(V)/(R_(1))`

Substituting the values we get

`I_(1)=(12)/(12)=1\ A`

Similarly, for current through R₂, R₃, R₄,

`I_(2)=(V)/(R_(2))=(12)/(3)=4\ A`

`I_(3)=(V)/(R_(3))=(12)/(5)=2.4\ A`

`I_(4)=(V)/(R_(4))=(12)/(4)=3\ A`

Now in Parallel Connection we have,

`I=I_(1)+I_(2)+I_(3)+I_(4)`

Substituting the values we get

`I =1+4+2.4+3=10.4\ Ampere`

Therefore,

The current through the battery is 10.4 Ampere.