Question

Three identical resistors are connected in parallel to a battery. If the current of 12. A flows from the battery, how much current flows through any one of the resistors

Answer 1

Answer:

4 A

Explanation:

We are given that

$R_1=R_2=R_3=4\Omega$

I=12 A

We have to find the current flowing through each resistor.

We know that in parallel combination current flowing through different resistors are different and potential difference across each resistor is same.

Formula :

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

Using the formula

$\frac{1}{R}=\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}$

$R=\frac{4}{3}\Omega$

$V=IR$

Substitute the values

$V=12\times \frac{4}{3}=16 V$

$I_1=\frac{V}{R_1}=\frac{16}{4}=4 A$

$I_1=I_2=I_3=4 A$

Hence, current flows through any one of the resistors is 4 A.