Question

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?

Answer 1

The ratio of final current to the original current is 0.67

Step-by-step explanation:

Given:

R₁ = 55 Ω

R₂ = 55 Ω

R₃ = 55 Ω

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

`(1)/(R) = (1)/(R_1) + (1)/(R_2)+ (1)/(R_3) \\\\(1)/(R) = (1)/(55) + (1)/(55) +(1)/(55) \\\\(1)/(R) = (3)/(55) \\`

`R = (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 =
`(V)/(R)`

I =
`(V)/(18.33)`

Case 2:

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

`(1)/(R)= (1)/(55)+ (1)/(55)\\ \\(1)/(R) =(2)/(55) \\\\R = (55)/(2) \\\\R = 27.5`

In this case, I =
`(V)/(27.5)`

Ratio of final current to original current =
`(V)/(27.5) :(V)/(18.33) \\\\`

`= (V)/(27.5) X (18.33)/(V) \\\\= 0.67`

The ratio of final current to the original current is 0.67