Answer :
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:
[tex]\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} \\[/tex]
[tex]R = \frac{55}{3} \\\\R = 18.33[/tex]
According to ohm's law:
V = IR
where,
V is the voltage
I is the current
In this case, I = [tex]\frac{V}{R}[/tex]
I = [tex]\frac{V}{18.33}[/tex]
Case 2:
When one of the device breaks then two resistors are left. The net resistance is:
[tex]\frac{1}{R}= \frac{1}{55}+ \frac{1}{55}\\ \\\frac{1}{R} =\frac{2}{55} \\\\R = \frac{55}{2} \\\\R = 27.5[/tex]
In this case, I = [tex]\frac{V}{27.5}[/tex]
Ratio of final current to original current = [tex]\frac{V}{27.5} :\frac{V}{18.33} \\\\[/tex]
[tex]= \frac{V}{27.5} X \frac{18.33}{V} \\\\= 0.67[/tex]
The ratio of final current to the original current is 0.67