Question

What is the resistance of a 1.00X10^2 -Ω , a 2.50-kΩ , and a 4.00-k Ω resistor connected in parallel?

Answer 1

Answer:

Therefore, the total Resistance:  R = 93.9 Ω

Step-by-step explanation:

Given

R₁ = 1 × 10²

R₂ = 2.5 kΩ = 2.5 × 10³ Ω

R₃ = 4  kΩ = 4 × 10³ Ω

Given that the given resisters are connected parallel, so using the formula to calculate the total Resistance R:

$\frac{1}{R}\:=\:\frac{1}{R_1}\:+\:\frac{1}{R_2}+\frac{1}{R_3}$

susbtituting R₁ = 1 × 10², R₂ = 2.5 × 10³ Ω, and R₃ = 4 × 10³ Ω

$\frac{1}{R}=\frac{1}{1\times \:10^2}+\frac{1}{2.5\times \:10^3}+\frac{1}{4\times \:10^3}$

Multiply by LCM of R, 100, 2500, and 4000:   20000 R

$\frac{1}{R}\cdot \:20000R=\frac{1}{1\cdot \:10^2}\cdot \:20000R+\frac{1}{2.5\cdot \:10^3}\cdot \:20000R+\frac{1}{4\cdot \:10^3}\cdot \:20000R$

simplify

$20000=213R$

Switch sides

$213R=20000$

Divide both sides by 213

$\frac{213R}{213}=\frac{20000}{213}$

Simplify

$R=\frac{20000}{213}$

$R = 93.9$ Ω

Therefore, the total Resistance:  R = 93.9 Ω