Réponse:
current in each branch is 6.4 A and 9.6 A.
Explanation:
Given that,
Current I = 16 A
First resistance R_{1}=8\text{\O}megaR
1
=8Ømega
Second resistance R_{2}=12\text{\O}megaR
2
=12Ømega
We know that,
The Current in 8 ohm,
I_{1} = \dfrac{R_{1}}{R_{1}+R_{2}}\times I_{total}I
1
=
R
1
+R
2
R
1
×I
total
I_{1} = \dfrac{8}{20}\times16I
1
=
20
8
×16
I_{1}= 6.4\ AI
1
=6.4 A
Now, The current in 12 ohm
I_{2}=16-6.4I
2
=16−6.4
I_{2}= 9.6\ AI
2
=9.6 A
Hence, The current in each branch is 6.4 A and 9.6 A.
