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Lina20 [59]
3 years ago
9

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?
Physics
1 answer:
Whitepunk [10]3 years ago
5 0

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.

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