Question

Two 4.0-Ω resistors are connected in parallel, and this combination is connected in series with 3.0 Ω. What is the effective resistance of this combination?

Answer 1

5.0 Ω

Step-by-step explanation:

`R_(1)` = 4.0 Ω,
`R_(2)` = 4.0 Ω,
`R_(3)` = 3.0 Ω

`(1)/(R_(parallel)) = (1)/(R_(1)) + (1)/(R_(2)) \\(1)/(R_(parallel)) = (1)/(4.0) + (1)/(4.0)\\(1)/(R_(parallel)) = (2)/(4.0)\\(1)/(R_(parallel)) = 0.5\\R_(parallel) = (0.5)^(-1) \\R_(parallel) = 2.0`

`R_(T) = R_(parallel) + R_(3) \\R_(T) = 2.0 + 3.0\\R_(T) = 5.0`

Therefore, the effective resistance of this combination is 5.0 Ω.

Answer 2

given,

R₁= 4 Ω

R₂ = 3 Ω

When two resistors are connected in series

R = R₁ + R₂

R = 4 + 3

R = 7 Ω

When two resistors are connected in series then their effective resistance is equal to 7 Ω .

When two resistors are connected in parallel.

`(1)/(R)=(1)/(R_1)+(1)/(R_2)`

`(1)/(R)=(1)/(3)+(1)/(4)`

`(1)/(R)=(7)/(12)`

`R = 1.714 \Omega`

Hence, the equivanet resistance in parallel is equal to
`R = 1.714 \Omega`