Electronics - Voltage and Current - Discussion
Discussion Forum : Voltage and Current - General Questions (Q.No. 2)
2.
If 60 J of energy are available for every 15 C of charge, what is the voltage?
Discussion:
162 comments Page 4 of 17.
Naveen said:
1 decade ago
v=w/q
Therefore v=60/15= 4v.
Therefore v=60/15= 4v.
Vijay said:
1 decade ago
i be the charge(15c)/time and p be the power 60j/watt so
p= vi
p= vi
Krishna said:
1 decade ago
q=c*v
q is charge,v is voltage,c is capacitence
60=15*v
Hence v=4volt
q is charge,v is voltage,c is capacitence
60=15*v
Hence v=4volt
Shashi said:
1 decade ago
charge=15 voltage=? energy=60
so voltage=energy/charge
i.e.60/15= 4volt.
so voltage=energy/charge
i.e.60/15= 4volt.
Ramraj said:
1 decade ago
Voltage= no of joules / coulomb.
60/15=4v.
60/15=4v.
Anju said:
1 decade ago
E=QV
V=E/Q
V=60/15
V=4
Where,
E=energy measured in joules
Q=charge measured in coulomb
v=voltage measured in volt
V=E/Q
V=60/15
V=4
Where,
E=energy measured in joules
Q=charge measured in coulomb
v=voltage measured in volt
Udayan Kamble said:
1 decade ago
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:
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:
Purnesh Nerale said:
1 decade ago
Thanks Friends.
Raghu said:
1 decade ago
Q=CV
SO,V=Q/C
*V=60/15
*V=4v
SO,V=Q/C
*V=60/15
*V=4v
Girish Bhat M said:
1 decade ago
P=VI
=> W/T=V(Q/T)
=> w=VQ
=> 60=V(15)
=. V=4v
=> W/T=V(Q/T)
=> w=VQ
=> 60=V(15)
=. V=4v
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