AcademyCurrent and Resistance
Academy
emf and Circuit Models
Level 1 - Physics topic page in Current and Resistance.
Principle
An emf source supplies energy per unit charge, and circuit models track how that energy is transferred around a closed loop.
Notation
\(\mathcal E\)
emf of a source
\(\mathrm{V}\)
\(W_{\mathrm{source}}\)
non-electrostatic work done by the source
\(\mathrm{J}\)
\(q\)
charge moved through the source
\(\mathrm{C}\)
\(r\)
internal resistance of a source
\(\mathrm{\Omega}\)
\(R\)
external load resistance
\(\mathrm{\Omega}\)
\(I\)
current in a simple loop
\(\mathrm{A}\)
\(V_{\mathrm{term}}\)
terminal voltage of the source
\(\mathrm{V}\)
Method
Derivation 1: Define emf as energy per charge
Inside a battery or generator, non-electrostatic processes move charge from lower potential to higher potential. The emf is the energy supplied per unit charge.
Energy supplied
\[W_{\mathrm{source}}=q\mathcal E\]
emf definition
\[\mathcal E=\frac{W_{\mathrm{source}}}{q}\]
Open circuit
\[I=0\Rightarrow V_{\mathrm{term}}=\mathcal E\]
Derivation 2: Include internal resistance
A real source can be modeled as an ideal emf in series with an internal resistance. When current leaves the positive terminal, some voltage is lost inside the source.
Internal drop
\[V_r=Ir\]
Terminal voltage
\[V_{\mathrm{term}}=\mathcal E-Ir\]
Load relation
\[V_{\mathrm{term}}=IR\]
Derivation 3: Simple one-loop model
For a source with internal resistance \(r\) connected to a load \(R\), the emf is used across both resistances.
Loop energy per charge
\[\mathcal E-IR-Ir=0\]
Total resistance
\[\mathcal E=I(R+r)\]
Loop current
\[I=\frac{\mathcal E}{R+r}\]
Rules
These are the compact source and loop relations.
emf definition
\[\mathcal E=\frac{W_{\mathrm{source}}}{q}\]
Terminal voltage
\[V_{\mathrm{term}}=\mathcal E-Ir\quad\text{discharging source}\]
Simple loop
\[I=\frac{\mathcal E}{R+r}\]
Open circuit
\[I=0\Rightarrow V_{\mathrm{term}}=\mathcal E\]
Examples
Question
A
\[9.0\,\mathrm{V}\]
battery with internal resistance \[0.50\,\Omega\]
drives a \[17.5\,\Omega\]
load. Find the current.Answer
\[I=\frac{\mathcal E}{R+r}=\frac{9.0}{17.5+0.50}=0.50\,\mathrm{A}\]
Checks
- emf is not a force; it is energy per charge.
- A real discharging source has terminal voltage below its emf.
- Internal resistance reduces current in a load circuit.
- Open-circuit terminal voltage equals emf only when no current flows.