3Work and power of current. Joule-Lenz's law

To use the formulae of work, power and efficiency of current source in problem solving;

If a charge Q moves between two points in a circuit that have a potential difference of V volts between them the energy gained (or lost) by the charge is given by the formula:
Electrical energy = Charge x Voltage
[Joules ]= [Coulombs x Volts = Amps x Time x Volts]
Electrical energy = I∙t∙V

Joule’s law formula: Q = I2 × R × t
where:
Q is the amount of heat, in Joules (J)
I is the electric current that flows through the wire, in amperes (A)
R is the value of the electrical resistance of the wire, in ohms (R)
t is the amount of time that current passes through the wire, in seconds (s).

Joule’s law can be established as the amount of heat (Q) generated in a wire with resistance (R), when a current (I) passes through it for a period of time (t).

This heat is directly proportional to:
The square of the current.
The resistance of wire.
The time the current flows through the wire.

What are the heating effects caused by electric current?

When a current flows in a wire, thermal energy is generated in it. The heating effects caused by electric current depend on three factors:

The resistance of the wire. Higher resistance produces more heat.
The time that current flows through the wire. The greater the time, the greater the heat generated.
The greater the current, the more heat generation.

Calculate the amount of energy supplied by a 4.5 V battery when:(a) a charge of 20 C passes through it(b) a current of 25 mA flows through is for
3 minutes

Energy = potential difference x
Q = 4.5 x 20 = 90 J(b) Energy = potential difference x Q =
p.d x current x time 
Therefore:Energy = 4.5 x 25x10-3 x 180 = 20.25 J