Question

In the following electrical network, the value of current I is:

पर्याय

• 2.7 A

• 4.8 A

• 5.9 A

• 6.6 A

Answer 1

6.6 A

Explanation:

Currents entering the main wire:

3 A, 4 A, 1.6 A, and 1.2 A

Total incoming = 3 + 4 + 1.6 + 1.2

= 9.8 A    ...(i)

Currents leaving the main wire:

0.8 A, 2.4 A, and I (unknown)

Total outgoing = 0.8 + 2.4 + I

= 3.2 + I    ...(ii)

Applying Kirchhoff’s current law:

Total current entering a junction = Total current leaving the junction

⇒ 9.8 = 3.2 + I    ...[From equation (i) and (ii)]

⇒ I = 9.8 − 3.2

= 6.6 A