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In the electricity topic, my physics book has defined Power as $P=IV$, $I$ being current and $V$ as voltage.
But then later in the Energy topic, power has been defined as $P=\frac{W}{t}=\frac{F.d}{t}$. The book also states that it is the same power we are talking about in both sections.
Therefore $IV=\frac{F.d}{t}$, But I don't see the logical link in this last equation. I think the definition from Electricity is a specific form of the definition from Energy.
Take $IV=\frac{dq}{dt}\int\vec{E}\cdot d\vec{l}=\frac{1}{dt}dq\left(\int\vec{E}\cdot\ d\vec{l}\right)=\frac{dw}{dt}$. You can think of it as the time rate of change of the work done by the electric field on a test charge, if you like.
Electrical Definition
Electric power, like mechanical power, is the rate of doing work, measured in watts, and represented by the letter $P$. The term wattage is used colloquially to mean "electric power in watts." The electric power in watts produced by an electric current $I$ consisting of a charge of $Q$ coulombs every $t$ seconds passing through an electric potential (voltage) difference of $V$ is
${\displaystyle P={\text{work done per unit time}}={\frac {VQ}{t}}=VI\,}$
Energy Definition
Power, as a function of time, is the rate at which work is done, (same definition as above). so it can be expressed by this equation:
${\displaystyle P(t)={\frac {W}{t}}}$
Because work is a force applied over a distance, this can be rewritten as:
${\displaystyle P(t)={\frac {W}{t}}={\frac {{\mathbf {F}}\cdot {\mathbf {d}}}{t}}}$
And with distance per unit time being a velocity, power can likewise be understood as:
${\displaystyle P(t)={\mathbf {F}}\cdot {\mathbf {v}}}$
Knowing from Newton's 2nd Law that force is mass times acceleration, the expression for power can also be written as:
${\displaystyle P(t)=m{\mathbf {a}}\cdot {\mathbf {v}}}$
Power will change over time as velocity changes due to acceleration. Knowing that acceleration is the time rate of change of velocity, this can then be written:
${\displaystyle P(t)=m{\mathbf {v}}\cdot {\frac {d{\mathbf {v}}}{dt}}}$
Comparing with the equation for kinetic energy:
${\displaystyle E_{\text{k}}={\tfrac {1}{2}}mv^{2}}$
It can be seen from the previous equation that power is mass times a velocity term times another velocity term divided by time. This shows how power is an amount of energy consumed per unit time.
The electrical power P(t)=I(t)·V(t) is a time dependent instantaneous power, which is defined as P=dW/dt, and P=W/t=F.d/t is an expression for the mean mechanical power when a mechanical work W was done in a time t. The instantaneous (time dependent) mechanical power P(t) has, in general, to be written as P(t)=F(t)·dd(t)/dt=F(t)·v(t). In general, you cannot equate the instantaneous electrical power to the mean mechanical power. If the instantaneous electrical power is equal to an instantaneous mechanical power, you should use the appropriate formula.
$$ P = IV = \frac{dq}{dt} \frac{dE}{dq} = \frac{dE}{dq} \frac{dq}{dt} $$
The last line is merely the chain rule, so
$$ P = \frac{dE}{dt} $$
You could imagine there is an amount of charge getting to the load per unit time $(I)$, and since each charge carries an amount of energy $(V)$, the total energy delivered per unit time will be $IV$.
