Question

three equal resistors of 2 ohm are connected in parallel and in turn connected in series with 10 ohm resistor and 16V battery. Calculate the total resistance in the circuit.

Answer 1

Explanation:

(a) Req =3+23×2=56=1.2Ω

(b) V=IR⇒I=RV

⇒I=1.212=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

Answer 2

Answer:

50/3 ohm

Explanation:

For parallel combination,

1/R = 1/R1 + 1/R2 + 1/R3

     = 1/2 + 1/2 + 1/2

     = 3/2

So, R = 2/3 ohm.

Now, its connected in series with 16 ohm

So, R = R1+R2

         = 2/3 + 16 = 50/3.

So, Total resistance = 50/3 ohm.