Capacitors

JEE Main level, single correct, medium

52 of 180

Question 52 of 180

Figure shows charge (q)(q) versus voltage (V)(V) graph for series and parallel combination of two given capacitors. The capacitances are:

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Answer: (D) 40μF40\mu F and 10μF10\mu F

Q=CVQ=CV
The slope of graph will give the equivalent capacitance of the combination.
From graph A:A: C1+C2=50010=50{C}_{1}+{C}_{2}=\dfrac{500}{10}=50 …(i)
From graph B:B: C1C2C1+C2=8010=8\dfrac{{C}_{1}{C}_{2}}{{C}_{1}+{C}_{2}}=\dfrac{80}{10}=8 …(ii)
From equation (i) & (ii)
C1C2=400{C}_{1}{C}_{2}=400 …(iii)
C1+400C1=50{C}_{1}+\dfrac{400}{{C}_{1}}=50 ⇒C12−50C1+400=0\Rightarrow {C}_{1}^{2}-50{C}_{1}+400=0
C1=50±2500−16002=50±302{C}_{1}=\dfrac{50\pm \sqrt{2500-1600}}{2}=\dfrac{50\pm 30}{2}
=10μF,40μF=10 \mu F, 40 \mu F

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