0
How much charge flows from each terminal of a 24.0 V battery when it is connected to a 7.90 F capacitor?
Wiki User
∙ 11y ago
For this problem, we use the equation: C=Q/V; where C=capacitor, Q=charge, V=volts
We are trying to find Q; the C and V are given, so we just need to plug the numbers into the equation.
C=7.90F
V=24.0 V
Q=?
7.90=Q/24.0
solve for Q
Wiki User
∙ 11y ago
Add your answer:
Earn +20 pts
A resistor R and a capacitor C are connected in series to a battery of terminal voltage V0 what equations relating the current you in the circuit and the charge Q on the capacitor de?
Current I=V0/R as per OHMs law: V=IR Charge on capacitor Q=CV=It
How much charge flows from each terminal of a 16.0 V battery when it is connected to a 8.80 F capacitor?
if the capacitor is not initially charged then it will receive during its charging an electric charge Q given by Q=V/C=16/8.8 = 1.82 microcoulomb
Can you charge a 12 volt battery with a 24 volt charger?
I can if I charge 2 of them at a time. Most 24 volt chargers also have a 12 volt setting. What is hard to find is a 32 volt charger. We can charge a 12V battery with 24V charger, by connecting two 12V batteries in series i. e. the negative terminal of the first battery is connected to the positive terminal of the second, now the charger positive terminal is connected to the first battery positive terminal and the negative terminal of the charger is connected to the negative terminal of the second battery.
An air-filled parallel-plate capacitor is connected to a battery and allowed to charge up Now a slab of dielectric material is placed between the plates of the capacitor while the capacitor is still?
the charge on the capacitor had increased.
How does a capacitor get charged?
a capacitor is used to store charge for longer period of time.in odder to understand that how a capacitor gets charged consider two parallel metal plates.these plates are neutral having equal amount of positive and negative charges.now connect these plates to the opposite terminals of a battery.the electrons present in the plate connected to the positive terminal of a battery will be attracted by the positive terminal leaving the plate positively charged.now these electrons are pushed by the battery to its negative terminal which repels these electrons to the plate connected to it making it negatively charged. now the battery is disconnected.so in this way the plates of a capacitor gets charged.as these plates have opposite charges stored on them force of attraction exist between them enabling a capacitor to store charge for longer periods of time.
Why capacitor is connected to earth in circuit?
when a capacitor is connected to earth the potential of capacitor becomes zero. as a result all the charge residing on the conductors of a capacitor passes away and the final charge on capacitor becomes zero
Describe the response of a capacitor when connected to and then disconnected from a battery?
A capacitor that is suddenly connected to a battery will charge to the battery voltage. The time to do this is dependent on the current capacity of the battery and wiring, and the capacitance of the capacitor. This represents an instantaneous short circuit, which lasts for a (usually) very short time - but damage could be done if there was no resistance. A charged capacitor that is suddenly disconnected from a battery will hold that voltage. The length of time it will hold is dependent on how much leakage current there is.
If you take out the battery in a capacitor before inserting a dielectric will there still be charge on the dielectric?
The charge in a capacitor is between the plates. The dielectric is only an insulator that allows the plates to be very close without touching and discharging the charge. There is no battery in a capacitor.
How does the charge vary with time in the R-C circuit?
well to get the answer first know the principle by which a capacitor(consider a capacitor without dielectric) gets charged .let a capacitor with plates p1 and p2 and resistor in parlell are connected to a DC source , when the switch is closed the circuit is in ON state and current starts flowing ,assume the flow of current as the movement of negative charges then the concept would be much clear. assume that p1 is connected with positive terminal of the battery and p2 with the negative terminal ,now as the switch is closed the negative charge on p1 is attracted by the positive terminal of the battery and is driven to the other plate p2 of capacitor.as this process continues charge seperation increases and potential difference starts getting developed and after a very long time the potential diff. across capacitor becomes equal to the applied voltage V. the maximum charge is Q0=CV and charge at any time is , Q(t)=Q0(1-e^(-t/T)) where T=time constant of the circuit
If a 0.40-uF capacitor is connected to a 9.0-V battery how much charge is on each plate of the capacitor?
The units of capacitance are called farads. A one farad capacitor is a capacitor with 1 volt potential difference with 1 coulomb of charge on the capacitor, C = Q/V or Q=CV So the charge held on your capacitor is Q = CV = 9Volts * 0.40*10-6Farads=3.6*10-6 Coulombs
When a battery is connected with a capacitor the potential difference will?
Charge the capacitor. Potential difference is a scientific term for what is more commonly called voltage. ANSWER: If big enough the battery will see a short initially and then proceed to charge the capacitor at a rate of 63% of the voltage in one time constant defined as RC For engineering purposes after 5 time the time constant the battery will and the capacitor zero potential different. The proper term should be virtual no difference.
What happens to the voltage across a capacitor as it discharges?
well to get the answer first know the principle by which a capacitor(consider a capacitor without dielectric) gets charged .let a capacitor with plates p1 and p2 and resistor in parlell are connected to a DC source , when the switch is closed the circuit is in ON state and current starts flowing ,assume the flow of current as the movement of negative charges then the concept would be much clear. assume that p1 is connected with positive terminal of the battery and p2 with the negative terminal ,now as the switch is closed the negative charge on p1 is attracted by the positive terminal of the battery and is driven to the other plate p2 of capacitor.as this process continues charge seperation increases and potential difference starts getting developed and after a very long time the potential diff. across capacitor becomes equal to the applied voltage V. during discharging first of all we short the battery , this mplies that we are short circuiting plates p1 and p2 so again charge redistribution takes place and after a longtime both the plates become electrically nuetral
