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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?

User Mudin
by
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2 Answers

3 votes

Answer:

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.

User Daxgirl
by
6.6k points
3 votes

Answer

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

User SMH
by
6.2k points