voltage as a difference in electric
potential energy.
Pātai: What is the third variable that we are
missing?
Resistance
Charge carries want to move around a circuit, and
in a conductor they can do this easily due to the mobile electrons.
In insulators they are not as able to move due to
the less mobile electrons.
Resistance is the idea of a friction that
the current encounters that reduces the current able to
flow, and causes energy to be lost/used as heat/light.
In reality, everything is a little bit of
a resistor.
Resistance has symbol
R in equations and has the unit Ohms
(\(\Omega\), the Greek letter
omega).
Resistance changes when circuit components
are added/removed or when a rheostat (variable resistor) is
altered
If supply voltage is constant, the current
will increase/reduce as the resistance changes.
Resistance & Heat
When current moves through a material with resistance the electrons
bump into other atoms. This causes energy to be transferred in
the form of vibrations (heat)!
The higher the resistance, the more heat
produced!
The higher the current, the more heat
produced!
Ohm’s Law
\[
\begin{aligned}
V &= IR \newline
voltage &= current \times resistance
\end{aligned}
\]
Voltage is measured in:
Current is measured in:
Resistance is measured in:
Pātai
The resistance of a light bulb is \(1.5k\Omega\). Calculate the current through
the bulb when it is connected across a \(12V\) power supply.
When \(9V\) is
applied to a resistor, \(0.03mA\) of
current flows through it. Calculate the resistance of the resistor.
How much voltage is required to produce \(180\mu A\) of current flowing through a
\(0.6M\Omega\) resistor?
Whakatika Tahi
The resistance of a light bulb is \(1.5k\Omega\). Calculate the current through
the bulb when it is connected across a \(12V\) power supply.
\[
\begin{aligned}
& V = IR \newline
& I = \frac{V}{R} \newline
& I = \frac{12}{1500} \newline
& I = 0.008A
\end{aligned}
\]
Whakatika Rua
When \(9V\) is applied to a
resistor, \(0.03mA\) of current flows
through it. Calculate the resistance of the resistor.
\[
\begin{aligned}
& V = IR \newline
& R = \frac{V}{I} \newline
& R = \frac{9}{0.00003} \newline
& R = 300000\Omega
\end{aligned}
\]
Whakatika Toru
How much voltage is required to produce \(180\mu A\) of current flowing through a
\(0.6M\Omega\) resistor?
\[
\begin{aligned}
& V = IR \newline
& V = (180 \times 10^{-6}) \times (0.6 \times 10^{6}) \newline
& V = 108V
\end{aligned}
\]
Power
Similar to Mechanics, power is defined as the amount of energy
transferred (work done) per second, and is measured in \(Js^{-1}\) or \(W\).
\[
\begin{aligned}
P &= IV
\end{aligned}
\]
The more power a component uses, the more energy is transformed
(e.g. into light, heat, etc.)
Pātai
Revisit the three pātai you did just before and calculate the power
dissapated (used) by the bulb and two resistors.