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Q.

A 3μF capacitor is charged to a 300 volt potential, and a 2μF capacitor is charged to a 200 volt potential. The capacitors are linked in parallel, with plates with opposite polarities connected together. After connecting the plates of the capacitor, the final potential difference between them is equal to:

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a

700V

b

240V

c

250V

d

260V

answer is D.

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Detailed Solution

Concept: According to the formula Q=CV, the quantity of charge that transfers into the plates is determined by the capacitance and the applied voltage, where Q is the charge in Coulombs, C is the capacitance in Farads, and V is the potential difference between the plates in volts.
We apply the formula of Capacitance Q =CV

V 1 = 300 V, C 2 = 2 μ F, V 2 = 200 V Q 1 = C 1 × V 1 Q 1 = (3 × 300) = 900 μ C Q 2 = C 2 × V 2 Q 2 = (2 × 200) = 400 μ C

⇒ V = Q 1 / C 1 = Q 2 / C 2

Now substitute the values we have,

900 − Q net / 3 = Q net + 400 / 2 ⇒ 2 (900 − Q net = 3 (Q net + 400) ⇒ (1800 − 2 Q n e t = (3 Q net + 1200) ⇒ 5 Q net = 600 ⇒ Q net = 600 / 5 = 120 μ C

So, the final potential difference is:

⇒ V = Q 1 / C 1 = Q 2 / C 2 ⇒ V = Q 1 − Q net / C 1 = Q net + Q 2 / C 2 ⇒ V = 900 − 120 / 3 = 780 / 3 = 260 volts .

Hence, the correct option is 4.

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A 3μF capacitor is charged to a 300 volt potential, and a 2μF capacitor is charged to a 200 volt potential. The capacitors are linked in parallel, with plates with opposite polarities connected together. After connecting the plates of the capacitor, the final potential difference between them is equal to: