Question

a current of 0.25 ampere flows in the main circuit now the resistance R is disconnected in and then connected across the 4 ohm resistance. current in the circuit is​

Answer 1

A current of 0.25 A flows in the main circuit. Now, the resistance R is disconnected and then connected across 4Ω resistance.

We have to find the current in the circuit now.

R and 2Ω are joined in parallel combination.

∴ R₁ = R × 2/(R + 2) = 2R/(R + 2)

now, R₁ , 4Ω and 3Ω resistors are in series combination.

∴ Req = R₁ + 4Ω + 3Ω = 7Ω + 2R/(R + 2)

From Ohm's law,

i = V/R

⇒0.25 = 2/Req

⇒0.25 × Req = 2

⇒Req = 7 + 2R/(R + 2) = 8

⇒R = 2Ω

now R is connected to 4Ω in the same way (i.e., parallel combination)

so, R₂ = 2 × 4/(2 + 4) = 4/3 Ω

now, R₂ , 2Ω and 3Ω are in series combination.

∴ Req = 4/3 + 2 + 3 = 19/3 Ω

from Ohm's law,

i' = V/R = 2/(19/3) = 6/19 A

Therefore the current in the circuit is now 6/19A.