You just multiply these two numbers. It's 5200J, or 5.2kJ
If a battery with a potential difference of 1.5 volts is placed across the plates, the maximum capacitor will have a charge of 36 V.
<h3>What possible variations are there in a 1.5 volt battery?</h3>
1 V is, by definition, a potential energy differential between two places equal to one joule for every coulomb of charge. Your query is resolved by that. Between the sites where that potential difference is measured, 1.5V denotes a potential energy differential of 1.5 joules per coulomb.
<h3>How do you determine the difference in potential energy?</h3>
ΔV=VB−VA=ΔPEq. By dividing the potential energy of a charge q that has been transported from point A to point B by the charge, we may define the potential difference between points A and B as VBVA. The joules per coulomb, sometimes known as volts (V) in honor of Alessandro Volta, are the units of potential difference.
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Answer:
120 watts
Explanation:
#1: 120 watts
#2: 667 watts
#3: 3 watts
#4: i forgot how to do this one
Answer:
voltage measured by the voltmeter = (E × RV)/(r + RV)
Explanation:
The circuit diagram for this description is presented in the attached image.
The internal resistance of an emf source is modelled to be in series with the source.
Therefore, the end product is a circuit with the battery in series connection with the internal resistance and resistance of the volunteer. The voltage picked up by the volunteer is the voltage across resistor RV.
Total resistance in the circuit = (r + RV) ohms (since both resistors are in parallel)
Current produced by the emf source = E/(Total resistance) = E/(r + RV)
The voltage across resistor RV = current flowing through this resistor × its resistance.
Since all the circuit elements are in series with each other, same current, E/(r + RV) flows through them all
Voltage across RV = voltage measured by the voltmeter = [E/(r + RV)] × RV = (E × RV)/(r + RV) = (E.RV)/(r + RV)