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Voltage, power and emf IB DP Physics Study Notes - 2025 Syllabus

Voltage, power and emf IB DP Physics Study Notes

Potential difference, current and resistance IB DP Physics Study Notes at  IITian Academy  focus on  specific topic and type of questions asked in actual exam. Study Notes focus on IB Physics syllabus with Students should understand

  • Ohm’s law
  • the ohmic and non-ohmic behaviour of electrical conductors, including the heating effect of resistors
  • electrical power P dissipated by a resistor as given by $P = IV = I^2R =\frac{ V^2}{ R}$

Standard level and higher level: There is no Standard level content
Additional higher level: 8 hours

IB DP Physics 2025 -Study Notes -All Topics

Ohm’s law

∙The German Ohm studied resistance of materials in the 1800s and in 1826 stated:
“Provided the temperature is kept constant, the resistance of very many materials is constant over a wide range of applied potential differences, and therefore the potential difference is proportional to the current.”
∙In formula form Ohm’s law looks like this:

   

FYI ∙

Ohm’s law applies to components with constant R.

Ohm’s law – ohmic and non-ohmic behavior

∙A material is considered ohmic if it behaves according to Ohm’s law. In other words the resistance stays constant as the voltage changes.

EXAMPLE:

Label appropriate V-I graphs with the following labels: ohmic, non-ohmic, R increasing, R decreasing, R constant.

SOLUTION:

• First label the resistance dependence.

• R = V/I so R is just the slope of the V vs. I graph.

• Ohm’s law states the R is constant. Thus only one graph is ohmic.
Since R is not constant the filament is non-ohmic.

EXAMPLE:

The graph shows the applied voltage V vs. resulting current I through a tungsten filament lamp.

Find R when \(I = 0.5 \, \text{mA}\) and \(1.5 \, \text{mA}\). Is this filament ohmic or non-ohmic?

▶️Answer/Explanation

SOLUTION:

• At \(0.5 \, \text{mA}\): \(V = 0.08 \, \text{V}\)
\[R = \frac{V}{I} = \frac{0.08}{0.5 \times 10^{-3}} = 160 \, \Omega.\]

• At \(1.5 \, \text{mA}\): \(V = 0.6 \, \text{V}\)
\[R = \frac{V}{I} = \frac{0.6}{1.5 \times 10^{-3}} = 400 \, \Omega.\]

Power dissipation

• Recall that power is the rate at which work is being done. Thus \(P = \frac{W}{t}\).

• From Topic 5.1 we learned that \(W = qV\).

• Thus
$P = \frac{W}{t} = \frac{qV}{t}$
$P = \left( \frac{q}{t} \right) V$
$P = IV.$

FYI
∙This power represents the energy per unit time delivered to, or consumed by, an electrical component having a current I and a potential difference V.

PRACTICE:

Use the definition of resistance \(R = \frac{V}{I}\) together with the one we just derived (\(P = VI\)) to derive the following two formulas:

(a) \(P = I^2R\)

(b) \(P = \frac{V^2}{R}\)

▶️Answer/Explanation

SOLUTION:

(a) From \(R = \frac{V}{I} \rightarrow V = IR\)
\(P = IV\)
\(= I(IR)\)
\(= I^2R\).

(b) From \(R = \frac{V}{I} \rightarrow I = \frac{V}{R}\)
\(P = IV\)
\(= \left( \frac{V}{R} \right) (V)\)
\(= \frac{V^2}{R}\).

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