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

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.

$\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}$

$\dfrac{1}{R}=\dfrac{1}{3}+\dfrac{1}{4}$

$\dfrac{1}{R}=\dfrac{7}{12}$

$R = 1.714 \Omega$

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

Answer 2

Answer:

5.0 Ω

Explanation:

$R_{1}$ = 4.0 Ω,     $R_{2}$ = 4.0 Ω,     $R_{3}$ = 3.0 Ω

$\frac{1}{R_{parallel}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} \\\frac{1}{R_{parallel}} = \frac{1}{4.0} + \frac{1}{4.0}\\\frac{1}{R_{parallel}} = \frac{2}{4.0}\\\frac{1}{R_{parallel}} = 0.5\\R_{parallel} = (0.5)^{-1} \\R_{parallel} = 2.0$

$R_{T} = R_{parallel} + R_{3} \\R_{T} = 2.0 + 3.0\\R_{T} = 5.0$

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