Two resistors in parallel
Two resistors in parallel
IP 2
7 marks
When two resistors of resistance \(R_1\) ohms and \(R_2\) ohms are connected in parallel, the total resistance \(R\) ohms satisfies \(\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}\), where \(R_1>0\) and \(R_2>0\).
- Show that \(R=\dfrac{R_1R_2}{R_1+R_2}\).[2]
- Make \(R_2\) the subject of the formula. Hence find the resistance \(R_2\) needed with \(R_1=6\) to give a total resistance of \(4\) ohms.[3]
- Explain why the total resistance is always less than \(R_1\).[2]
Solution:
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Answer:(a) Shown (b) \(R_2=\dfrac{RR_1}{R_1-R}\), \(12\) ohms (c) \(R=R_1\times\dfrac{R_2}{R_1+R_2}\) and the fraction is less than \(1\)
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