Electronics - Voltage and Current - Discussion

VOLTS x COULOMBS = JOULES
ANSWER = 60/15 = 4
It takes energy to push some charge against the voltage pressure

Whenever a certain amount of charge is pushed through an electrical resistance, some electrical energy is lost from the circuit and heat is created. A certain amount of energy flows into the "frictional" resistor every second, and a certain amount of heat energy flows back out again. If we increase the voltage, then for the same hunk of charge being pushed through, more energy flows into the resistor and gets converted to heat. If we increase the hunk of charge, same thing: more heat flows out per second. Here's how to write this:

Chetan Kumar and Jaimin Patel Gujarat you both are wrong why because
you said that Energy=1/2CV^2
It's Correct for Electrical Energy stored in a Capacitor Only.
Then
U=QV/2=CV^2/2

Where U=Potential Energy
Q=Coulombs(C)
V=Volts(V)
C=Capacitance in frads.

But In the above problem we should use the formula that
Energy(Q)=Charge(C)*Voltage(V)
Q=CV
V=Q/C; Where V stands for Voltage

Thanks to all of you and especially for Vinod who told that Chetan was wrong.If we know the correct answer then only we can say that to anyone But to say it's wrong we should have guts that type of daring i got from Vinod.Thank you Vinod.

The Unit For The Measurement Of Charge Is Coulomb Which is Equal To 6.28*10^18 Electrons/Second.

The Unit For The Measurement Of Current Is Ampere Which Is Equal To 1 Coulomb/Second.

Therefore, Current = 15C/Second.

Electrical Energy = Power*Time = 60J (KWH).

Power = Current*Voltage.

Therefore, Electrical Energy = Current*Voltage*Time.

Therefore, 60J = 15C*Voltage*1 Second.

Therefore, Voltage = 60J/15C*1.

Therefore, Voltage = 4V*1.

Therefore, Voltage = 4V.

Let V=voltage to be found, I=current passing through the circuit in time 't' & Q=charge in coulombs. Then, electrical power P & current I are given by

P=V*I.....(1)
&
I=dQ/dt...(2)

so, the electrical energy E is

E=V*Q.....(3)

Since E=60 J(Given) & Q=15 C(Given),

V=E/Q=(60/15) V =4 V

Let V=voltage to be found, I=current passing through the circuit in time 't' & Q=charge in coulombs. Then, electrical power P & current I are given by,

P = V*I.....(1).
&
I = dQ/dt...(2).

So, the electrical energy E is:

E = V*Q.....(3).

Since E = 60 J(Given) & Q = 15 C(Given),

V = E/Q = (60/15). V = 4 V.

Hey
Here problem is about a single charge . if suppose +Ve charge has 15C charge and has energy 60 J than what is the value of potential generated by it.
we can not treat it as capacitor because in capacitor we treat the potential difference applied between plates.so kindly have it simple case. E = q . V = = 4 volt

Here problem is about a single charge . if suppose +Ve charge has 15C charge and has energy 60 J than what is the value of potential generated by it.
we can not treat it as capacitor because in capacitor we treat the potential difference applied between plates.so kindly have it simple case. E = q . V = = 4 volt

Joules are volt-coulombs; that is, when you put one coulomb of electrons through a potential difference of 1 volt, you do one joule of work on the system. So you find the number of joules of work (or energy put in to the system) by multiplying the number of coulombs by the number of volts.

The potential difference or voltage, V across two points is defined as energy, E dissipated or transferred by coulomb of charge, Q that moves through the two points.
Therefore:

Potential difference = Electrical energy dissipated
Charge

V= E /Q

V= Voltage
E= Energy
Q= Charge

E = 1/2 * C * V^2
With the help of this Einsten Eq. the Ans. Should be 8V.
but this is not a option, so we can go through to this way

I= Q/T &........1
P=V*I ,........2
On putting the value of I in equation 2.
P=V*Q/T
E=P*T
E=V*Q/T*T
E=V*Q
V=E/Q
V=60J/15Q
V=4V.