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JEE Main 2019 (10 Apr, Shift 1) · Combination of Capacitors

Combination of Capacitors question from JEE Main 2019 (10 Apr, Shift 1)

Q75JEE Main 2019 · 10 APR · SHIFT 1Medium

Combination of Capacitors

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

Figure

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Answer: (a)

  1. Larger slope is parallel: Cp=50010=50 μFC_p = \frac{500}{10} = 50\ \mu\text{F}; smaller is series: Cs=8010=8 μFC_s = \frac{80}{10} = 8\ \mu\text{F}.
  2. C1+C2=50C_1 + C_2 = 50 and C1C2C1+C2=8⇒C1C2=400\frac{C_1C_2}{C_1+C_2} = 8 \Rightarrow C_1C_2 = 400.
  3. So C1,C2C_1, C_2 are roots of x2−50x+400=0x^2 - 50x + 400 = 0: 4040 and 1010.
  4. So the answer is (a) 40 μF40\ \mu\text{F} and 10 μF10\ \mu\text{F}.
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