a wire of resistance 80 ohm is cut into 8 equal pieces. these pieces are joined in parallel. find the net resistance of its combination
.
Plzzz answer fast
Answers
Answer:
0.8 ohm.
this is the answer.......
Answer:
The correct answer is 8.89 ohm.
Explanation:
The expression for the equivalent resistance in the parallel combination is as follows;
\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}
R
1
=
R
1
1
+
R
2
1
+
R
3
1
Here, R{1},R_{2},R_{3}R1,R
2
,R
3
are the resistances.
It is given in the problem that a wire whose resistance is 80 ohm is cut into three pieces of equal length.
The value of each resistance is equal to \frac{80}{3}
3
80
.
Put R{1}=R_{2}=R_{3}=\frac{80}{3}R1=R
2
=R
3
=
3
80
.
\frac{1}{R}=\frac{3}{80}+\frac{3}{80}+\frac{3}{80}
R
1
=
80
3
+
80
3
+
80
3
R=\frac{80}{9}R=
9
80
R=8.89 ohmR=8.89ohm
Therefore, resistance of the given expression is 8.89 ohm.
Buddy I am not sure if this is what you wanted
If anything else I could help you with I would be happy to