Electrostatics 5 Question 4

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

(Main 2019, 10 April I)

(a) 60μF and 40μF

(b) 50μF and 30μF

(c) 20μF and 30μF

(d) 40μF and 10μF

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Answer:

Correct Answer: 4. (d)

Solution:

  1. In the given figure, Slope of OA> Slope of OB

Since, we know that, net capacitance of parallel combination > net capacitance of series combination

Parallel combination’s capacitance,

CP=C1+C2=500μC10 V=50μF(i)

Series combination’s capacitance,

CS=C1C2C1+C2=80μC10 V=8μF(ii)

or

C1C2=8×(C1+C2)=8×50μF(iii)=400μF [using Eq. (i)] 

From Eqs. (i) and (iii), we get

C1=50C2 and C1C2=400C2(50C2)=40050C2C22=400 or C2250C2+400=0C2=+50±250016002=+50±302

C2=+40μF or +10μF

Also, C1=50C2C1=+10μF or +40μF

Hence, capacitance of two given capacitors is 10μF and 40μF.



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