To resistances when connected in parallel give resultant value of 2 ohm; when connected in series the value becomes 9 ohm. Calculate the value of each resistance.
Answers
Answer:
Explanation:
The resistance of the resistor are 3 ohm and 6 ohm.
Given:
The resultant resistance when two resistances are connected in series = 2 ohms
The resultant resistance when two resistances are in parallel connected = 9 ohms
To calculate:
The value of each resistance = ?
Solution:
Resistor value when resistors are connected in parallel,
\frac{1}{R_{p}}=\frac{1}{R 1}+\frac{1}{R 2}
\Rightarrow \frac{1}{2}=\frac{1}{R 1}+\frac{1}{R 2} \rightarrow(1)
Resistances in series,
R_{s}=R 1+R 2
We know that, resistance is 9 ohm in series,
\Rightarrow 9=R 1+R 2
\Rightarrow R 2=9-R 1
On substituting in equation (1), we get,
\frac{1}{2}=\frac{1}{R 1}+\frac{1}{(9-R 1)}
\frac{1}{2}=\frac{(9-R 1)+R 1}{R 1(9-R 1)}
9 R 1-R 1^{2}=18-2 R 1+2 R 1
\Rightarrow R 1^{2}-9 R 1+18=0
\Rightarrow R 1^{2}-6 R 1-3 R 1+18=0
\Rightarrow(R 1-6)(R 1-3)=0
\therefore R 1=6,3
When R1 is 3, R2 is 6
When R1 is 6, R2 is 3