Question

In the circuit shown below current I flowing through 6 ohm resistence and voltage drop V across 3 ohm resistence respectively are? ??



Please answer fast I will mark brainiest. .....

Answer 1

Solution:

$\boxed{\sf{I=\frac{10}{3+2} = \frac{10}{5} = 2\:A}}$

From figure 3,

We can see that as in both branches equal resistance of $\sf{4\:\pi}$ is there. So current through them will be equal:

$\implies$ $\boxed{\sf{\frac{I}{2} = 1\:A}}$

From fig.2 and fig.1, we can see that:

$\implies$ $\sf{\frac{I}{2} = I_{1} + I_{2}}$

$\implies$ $\sf{V_{AB} = V_{CD}}$

Therefore:

$\implies$ $\sf{I_{1} \times 6 = I_{2} \times 3}$

$\implies$ $\sf{I_{2} = 2I_{1} = 2I_{0}}$ (∴ $\sf{I_{1}=I_{0}}$)

$\implies$ $\sf{\frac{I}{2} = I_{0} + 2I_{0}}$

$\implies$ $\sf{\frac{I}{2} = 3I_{0}}$

So:

$\implies$ $\boxed{\sf{I_{0} = \frac{I}{6} = \frac{2}{6} = \frac{1}{3}\:A}}$

Therefore:

$\implies$ $\sf{V_{0} = V_{AB} = V_{CD}}$

$\implies$ $\sf{V_{0} = I_{0} \times 6}$

$\implies$ $\sf{V_{0} = \frac{1}{3} \times 6}$

$\implies$ $\sf{V_{0} = 2\:V}$

________________

Answered by: Niki Swar, Goa❤️

Answer 2

Explanation:

R1=3ohm R2=4ohm, R3=6ohm

As the resistors are connected in parallel

1/Rtotal=1/3+1/4+1/6

1/Rtotal=(4+3+2)/12

1/Rtotal=9/12

Rtotal=12/9

Rtotal=4/3 ohm

Total current=Voltage /Rtotal

Itotal=2/(4/3)

Itotal=6/4

Itotal=3/2 Ampere

Current through R1=V/R1=2/3 Ampere

Current through R1=V/R3=2/6=1/3 Ampere