Physics | VITEEE 2017 | Discussion Page

Two capacitors C1and C₂ in a circuit are joined as shown in figure. The potentials of points A and B are V₁ and V₂ respectively. Then the potential of point D will be

$\frac{V_{1}+V_{2}}{2}$

$\frac{C_{2}V_{1}+C_{1}V_{2}}{C_{1}+C_{2}}$

$\frac{C_{1}V_{1}+C_{2}V_{2}}{C_{1}+C_{2}}$

$\frac{C_{2}V_{1}+C_{1}V_{2}}{C_{1}+C_{2}}$

Answer:

Option C

Explanation:

Consider the potential at D be 'V'.

Potential drop across C₁ is (V - V₁) and C₂ is (V - V)

.'. q₁ = C₁(V - V₁), q₂ = C₂(V₂ - V)

As q₁ = q₂ [capacitors are in series]

.'. C₁(V - V₁) = C₂(V₂ - V)

V = $\frac{C_{1}V_{1}+C_{2}V_{2}}{C_{1}+C_{2}}$

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Answer:

Explanation: