Question

A 2uF capacitor is charged to 200 volt and then the battery is disconnected. When it is connected in parallel to another uncharged capacitor, the potential difference between the plates of both is 40 volt. The capacitance of the other capacitor is :

A) 2 uF
B) 8 uF
C) 4uF
D) 16 uF

Answer 1

By applying the conservation of charge to the parallel connection of a charged 2uF capacitor and an uncharged capacitor, the calculated capacitance of the uncharged capacitor is found to be 8uF.

Step-by-step explanation:

The question involves finding the capacitance of an uncharged capacitor when it is connected in parallel with a charged capacitor. Initially, a 2uF capacitor is charged to 200V, and once the battery is disconnected, this charged capacitor is connected in parallel with another uncharged capacitor, leading to a potential difference of 40V across both capacitors.

To find the capacitance of the uncharged capacitor, we can use the conservation of charge. Charge before connecting (Qinitial) is equal to the charge after connecting (Qfinal), as no external charge is added or removed from the system. Given that:

• Qinitial = C1 × V1 = 2uF × 200V = 400μC
• Qfinal = (C1 + C2) × Vfinal = (2uF + C2) × 40V

Setting Qinitial equal to Qfinal:

1. 400μC = (2uF + C2) × 40V
2. 400μC = 80μC + 40VC2
3. 320μC = 40VC2
4. C2 = 320μC / 40V = 8uF

So, the capacitance of the uncharged capacitor is 8uF. Therefore, the correct answer is B) 8 uF.