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
[tex]I =\dfrac{V}{R}[/tex]
So current through R₁
[tex]I_{1}=\dfrac{V}{R_{1}}[/tex]
Substituting the values we get
[tex]I_{1}=\dfrac{12}{12}=1\ A[/tex]
Similarly, for current through R₂, R₃, R₄,
[tex]I_{2}=\dfrac{V}{R_{2}}=\dfrac{12}{3}=4\ A[/tex]
[tex]I_{3}=\dfrac{V}{R_{3}}=\dfrac{12}{5}=2.4\ A[/tex]
[tex]I_{4}=\dfrac{V}{R_{4}}=\dfrac{12}{4}=3\ A[/tex]
Now in Parallel Connection we have,
[tex]I=I_{1}+I_{2}+I_{3}+I_{4}[/tex]
Substituting the values we get
[tex]I =1+4+2.4+3=10.4\ Ampere[/tex]
Therefore,
The current through the battery is 10.4 Ampere.
