Question

A current of 16 amperes divides between two branches in parallel of resistances 8 ohms and 12 ohms respectively. The current in each branch is

Answer 1

i is inversely proportional to resistance....

Answer 2

Answer:

The current in each branch is 6.4 A and 9.6 A.

Explanation:

Given that,

Current I = 16 A

First resistance $R_{1}=8\Omega$

Second resistance $R_{2}=12\Omega$

We know that,

The Current in 8 ohm,

$I_{1} = \dfrac{R_{1}}{R_{1}+R_{2}}\times I_{total}$

$I_{1} = \dfrac{8}{20}\times16$

$I_{1}= 6.4\ A$

Now, The current in 12 ohm

$I_{2}=16-6.4$

$I_{2}= 9.6\ A$

Hence, The current in each branch is 6.4 A and 9.6 A.