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Capacitor question

2019 · 10 Apr · Shift 1 · Q62

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JEE Main
Physics
Capacitor
2019 · 10 Apr · Shift 1 · Q62

Capacitor question

2019 · 10 Apr · Shift 1 · Q62

JEE MainPhysicsCapacitorMCQ+4 / −1

Figure shows charge (q) versus voltage (V) graph for series and parallel combination of two given capacitors. The capacitances are : JEE Main 2019 (Online) 10th April Morning Slot Physics - Capacitor Question 116 English

JEE Main 2019 (Online) 10th April Morning Slot Physics - Capacitor Question 116 English

A
40 $\mu$ F and 10 $\mu$ F
B
60 $\mu$ F and 40 $\mu$ F
C
20 $\mu$ F and 30 $\mu$ F
D
50 $\mu$ F and 30 $\mu$ F

View written solutionFree

Correct answer: A

For a capacitor, the relation between charge and voltage is $q=CV$ So, in a $q$ vs $V$ graph, the slope equals the capacitance: $\text{slope}=\frac{q}{V}=C$
Let the two capacitors be $C_1$ and $C_2$ .

In parallel combination: $C_p=C_1+C_2$
In series combination: $C_s=\frac{C_1C_2}{C_1+C_2}$

From the graph, the steeper line corresponds to the parallel combination and the less steep line corresponds to the series combination.

From the given graph, the capacitances obtained from slopes are: $C_p=50\,\mu F, \qquad C_s=8\,\mu F$

Now use $C_1+C_2=50$ and $\frac{C_1C_2}{C_1+C_2}=8$

Since $C_1+C_2=50$ , we get $C_1C_2=8\times 50=400$

So $C_1$ and $C_2$ satisfy $x^2-50x+400=0$

Solve the quadratic: $x^2-50x+400=0$ $x=\frac{50\pm\sqrt{2500-1600}}{2}$ $x=\frac{50\pm\sqrt{900}}{2}$ $x=\frac{50\pm 30}{2}$

Thus, $x=40\,\mu F \quad \text{or} \quad 10\,\mu F$

Therefore, the two capacitances are $40\,\mu F \text{ and } 10\,\mu F$

So the correct option is A.

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