Question

Refer the attached picture. ​

Answer 1

Answer is in Attachment

Thank you

Answer 2

Given : A circuit is given in which different resistors are joined as shown in figure.

To find : The current flowing through the resistor R.

solution : 1Ω, 2Ω, 8Ω, 2Ω and 4Ω are formed a balanced Wheatstone bridge. so 8Ω resistor must be removed.

so, equivalent resistance of these :

1/R₁ = 1/(1 +2) + 1/(2 + 4) = 1/3 + 1/6

⇒R₁ = 6 × 3/(3 +6) = 2 Ω

now R₁ , 4Ω, 10Ω, 6Ω and 12Ω are also formed a balanced Wheatstone bridge. so, 10Ω resistor must be removed.

so, equivalent resistance of these :

1/R₂ = 1/(2 + 4) + 1/(6 + 12)

⇒1/R₂ = 1/6 + 1/18 = 3/18 + 1/18 = 4/18

⇒R₂ = 4.5 Ω

now R₂ and R are joined in parallel combination

so, Req = R₂ + R = 2 + 4.5 = 6.5 Ω

current, I = V/R

= 6.5/6.5 = 1 A

Therefore the current flowing through the resistor R is 1 Ampere.