Question
Two 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.
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Answer
Two resistance when connected in series, resultant value is 9 ohms. Two resistance when connected in parallel, resultant values is 2 ohms. Let the two resistance be R1 and R2 If connected in series, then 9 = R1 + R2 R1 = 9 - R2 If connected in parallel, then
$\frac{1}{2}=\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}$
From aboves equations we get that$\frac{1}{2}=\frac{(\text{R}_1+\text{R}_2)}{\text{R}_1\text{R}_2}$
$\frac{1}{2}=\frac{9}{(9-\text{R}_2)\text{R}_2}$
$9\text{R}_2-\text{R}_2{^2}=18$
$\text{R}_2{^2}-9\text{R}_2+18=0$
$(\text{R}_2-6)(\text{R}_2-3)=0$
$\text{R}_2=6,3$
So if R2 6 ohms, then R1 = 9 - 6 = 3 ohms. If R2 = 3 ohms, then R1 = 9 - 3 = 6 ohms.Need a full question paper?
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For the circuit shown in the diagram below:
What is the value of :
Current through $6 Ω$ resistor?
Potential difference across $12 Ω$ resistor?
→What is the value of :
Current through $6 Ω$ resistor?
Potential difference across $12 Ω$ resistor?