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

Answer:

Therefore,

The current through the battery is 10.4 Ampere.

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

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

So current through R₁

$I_{1}=\dfrac{V}{R_{1}}$

Substituting the values we get

$I_{1}=\dfrac{12}{12}=1\ A$

Similarly, for current through R₂, R₃, R₄,

$I_{2}=\dfrac{V}{R_{2}}=\dfrac{12}{3}=4\ A$

$I_{3}=\dfrac{V}{R_{3}}=\dfrac{12}{5}=2.4\ A$

$I_{4}=\dfrac{V}{R_{4}}=\dfrac{12}{4}=3\ A$

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.