Capacitors and switches problem

In summary, when the switch is closed, the potential difference across both capacitors remains the same as before, which is 60pC. This is due to the fact that two capacitors connected in series hold the same charge and there can be no current flow since one end of each capacitor is open. The given answer in the book, stating that the potential difference becomes equal to the average of V1 and V2, is incorrect as the resulting capacitance is 1.2 pF, resulting in a potential difference of 50V.
  • #1
kihr
102
0

Homework Statement


Two capacitors, one with C1=2pF and V1=30V, and the other with C2=3pF and V2=20V are connected to each other through a switch (positive end of one connected to the negative end of the other). The other end of each capacitor is kept open. What happens to V1 and V2when the switch is closed?


Homework Equations




The Attempt at a Solution



Both the capacitors have the same charge, i.e. 60pC. When the switch is closed there will be no charge flowing between the two capacitors as two capacitors connected in series hold the same charge. Moreover, since one end of each capacitor is open, there can be no current flow in any case. Hence there will be no change in V1 and V2 when the switch is closed.

The answer gven in the book mentions that when the switch is closed the potential difference across C1 and C2 will become equal to the average of V1 and V2, i.e. = 25 V. I am not able to reconcile with this logic. Hence would appreciate some views of the Forum members.
 
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  • #2


You are right, the voltage remains the same across both capacitors.

The situation is the same as if you had two uncharged capacitors, 2 pF and 3 pF, connected in series, and then charged to 60 pC. The resultant capacitance is 1.2 pF, so the overall potential difference is V=q/C=50 V. V is shared between the two capacitors: V=V1+V2. V1=q/C1=30 V and V2=20 V. The book states nonsense.

ehild
 

1. What is the purpose of a capacitor in a circuit?

A capacitor is an electronic component that stores electrical energy in the form of an electric field. It is used in circuits to temporarily store and release energy, filter out unwanted noise, and regulate voltage levels.

2. How does a capacitor work?

A capacitor is made up of two conductive plates separated by an insulating material called a dielectric. When a voltage is applied across the plates, an electric field is created, causing one plate to become positively charged and the other negatively charged. This creates a potential difference, or voltage, between the plates.

3. What is the role of a switch in a circuit?

A switch is an electronic component that allows or blocks the flow of electricity in a circuit. It is used to control the on/off state of a circuit or to redirect the flow of electricity to different components.

4. How does a switch function in a circuit with a capacitor?

In a circuit with a capacitor, a switch can be used to control the charging and discharging of the capacitor. When the switch is closed, the capacitor charges as it allows current to flow through it. When the switch is opened, the capacitor discharges as it releases the stored energy.

5. Can a switch and a capacitor be used together to create a time delay in a circuit?

Yes, a switch and a capacitor can be used together to create a time delay in a circuit. By controlling the charging and discharging of the capacitor, the switch can be used to introduce a delay in the flow of current. This is commonly used in electronic timers and time delay circuits.

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