Physics, 10.04.2020 01:03 Ruchen

Hey circuit includes three resistors in parallel, each with a resistance of 55 ohm's. If one of the devices breaks, what is the ratio of the final current to the original current?

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Answer from: MIa2020

The ratio of final current to the original current is 0.67

Explanation:

Given:

R₁ = 55 Ω

R₂ = 55 Ω

R₃ = 55 Ω

When the resistors are connected in parallel then the net resistance, R is:

$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}+ \frac{1}{R_3} \\\\\frac{1}{R} = \frac{1}{55} + \frac{1}{55} +\frac{1}{55} \\\\\frac{1}{R} = \frac{3}{55} \\$

$R = \frac{55}{3} \\\\R = 18.33$

According to ohm's law:

V = IR

where,

V is the voltage

I is the current

In this case, I = $\frac{V}{R}$

I = $\frac{V}{18.33}$

Case 2:

When one of the device breaks then two resistors are left. The net resistance is:

$\frac{1}{R}= \frac{1}{55}+ \frac{1}{55}\\ \\\frac{1}{R} =\frac{2}{55} \\\\R = \frac{55}{2} \\\\R = 27.5$

In this case, I = $\frac{V}{27.5}$

Ratio of final current to original current = $\frac{V}{27.5} :\frac{V}{18.33} \\\\$

$= \frac{V}{27.5} X \frac{18.33}{V} \\\\= 0.67$

The ratio of final current to the original current is 0.67

Answer from: Quest

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Answer from: Quest

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