A circuit includes three resistors in parallel, each with a resistance of 55 . If one of the devices breaks, what is the ratio of the final current to the original current?
Answers
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:
\begin{gathered}\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} \\\end{gathered}
R
1
=
R
1
1
+
R
2
1
+
R
3
1
R
1
=
55
1
+
55
1
+
55
1
R
1
=
55
3
\begin{gathered}R = \frac{55}{3} \\\\R = 18.33\end{gathered}
R=
3
55
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}
R
V
I = \frac{V}{18.33}
18.33
V
Case 2:
When one of the device breaks then two resistors are left. The net resistance is:
\begin{gathered}\frac{1}{R}= \frac{1}{55}+ \frac{1}{55}\\ \\\frac{1}{R} =\frac{2}{55} \\\\R = \frac{55}{2} \\\\R = 27.5\end{gathered}
R
1
=
55
1
+
55
1
R
1
=
55
2
R=
2
55
R=27.5
In this case, I = \frac{V}{27.5}
27.5
V
Ratio of final current to original current = \begin{gathered}\frac{V}{27.5} :\frac{V}{18.33} \\\\\end{gathered}
27.5
V
:
18.33
V
\begin{gathered}= \frac{V}{27.5} X \frac{18.33}{V} \\\\= 0.67\end{gathered}
=
27.5
V
X
V
18.33
=0.67
The ratio of final current to the original current is 0.67