For a given potential difference V how would you connect two capacitors, to obtain greater stored charge :

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Answer 1

Since $q = C V$ & capacitance of two capacitor is more in parallel combination as compared to series combinations i.e., $C_{P} > C_{S}$ .

Answer 2

Since

q = C V

& capacitance of two capacitor is more in parallel combination as compared to series combinations i.e.,

C_{P} > C_{S}

.

For a given potential difference V how would you connect two capacitors, to obtain greater stored charge :

series

parallel

series and parallel combined

cannot say

Solution

Since $q = C V$ & capacitance of two capacitor is more in parallel combination as compared to series combinations i.e., $C_{P} > C_{S}$ .

Patterns of problems

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For a given potential difference V how would you connect two capacitors, to obtain greater stored charge :

series

parallel

series and parallel combined

cannot say

Since q=CV & capacitance of two capacitor is more in parallel combination as compared to series combinations i.e., CP​>CS​.

Patterns of problems

Was this answer helpful?

Similar questions

Two capacitors, C1​=2μF and C2​=8μF are connected in series across a 300V source. Then

A capacitor of capacitance C is charged to a potential difference V from a cell and then disconnected from it. A charge + Q is now given to its positive plate. The potential difference across the capacitor is now :

Two capacitors of capacitance C1​ and C2​ are connected in parallel across a battery. If Q1​ and Q2​ respectively be the charges on the capacitors, then Q2​Q1​​ will be equal to :

A capacitor C1​ is charged to a potential difference V. The charging battery is then removed and the capacitor is connected to an uncharged capacitor C2​. The potential difference across the combination is

More From Chapter

Electrostatic Potential and Capacitance

Shortcuts & Tips

Diagram set

Mindmap

Cheatsheets

Common misconceptions

Problem solving tips

Memorization tricks

Practice more questions

8 Qs

2 Qs

3 Qs

3 Qs

1 Qs

51 Qs

Since $q = C V$ & capacitance of two capacitor is more in parallel combination as compared to series combinations i.e., $C_{P} > C_{S}$ .

Two capacitors, $C_{1} = 2 μ F$ and $C_{2} = 8 μ F$ are connected in series across a $300 V$ source. Then

Two capacitors of capacitance $C_{1}$ and $C_{2}$ are connected in parallel across a battery. If $Q_{1}$ and $Q_{2}$ respectively be the charges on the capacitors, then $\frac{Q _{1}}{Q _{2}}$ will be equal to :

A capacitor $C_{1}$ is charged to a potential difference $V$ . The charging battery is then removed and the capacitor is connected to an uncharged capacitor $C_{2}$ . The potential difference across the combination is