Question

Q Two conductors of resistances
3 ohm and 4 ohm are connected in series across a battery of 12V. find the Current flowing throught the circiuit.​

Answer 1

Answer:

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Explanation:

Req =

3+2

3×2

=

5

6

=1.2Ω

(b) V=IR⇒I=

R

V

⇒I=

1.2

12

=10A

(c) 3(10−i)=2i (∵ Potential drop in parallel combination are same)

⇒30−3i=2i⇒30=5i⇒i=6A

Current in 2Ω resistor is =i=6A

Current in 3Ω resistor is =10−i=4A