EMF

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Table of Contents

• EMF
  • Key Definitions
  • EMF
    • $\epsilon =\int {\overrightarrow{f}}_s\cdot d\overrightarrow{l}$
      • ${\overrightarrow{f}}_s=\frac{{\overrightarrow{F}}_s}{q}\left[\frac{N}{C}\right]$
    • Power in a Battery
      • $P=IV=I\epsilon$
    • EMF in a Battery
      • ${\overrightarrow{F}}_{net}=q(\overrightarrow{E}+{\overrightarrow{f}}_s)$
    • EMF in a Circuit
      • $\epsilon =I(R+r)$
        • $I=\frac{\epsilon }{R+r}$
      • $V_{battery}=IR=\epsilon -Ir$
  • Circuits
    • Open Circuits
      • $I=0$
      • $V_{battery}=\epsilon -0=\epsilon$
    • Short Circuit
      • $I_{max}=\frac{\epsilon }{r}$
      • $V=\epsilon -\epsilon =0$
      • $P_{max}=\frac{{ϵ}^2}{r}$

#UCLA #Y1Q3 #Physics1B

EMF

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Key Definitions

DC Circuit - with one-directional flow of charge

Electromotive Force (EMF) - the extra force in a battery (usually from chemical energy) required to push

• represented as the “battery” symbol in a circuit

Ideal Battery - a battery with an internal resistance of 0

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EMF

Defined as the potential:

$\epsilon =\int {\overrightarrow{f}}_s\cdot d\overrightarrow{l}$

Where

${\overrightarrow{f}}_s=\frac{{\overrightarrow{F}}_s}{q}\left[\frac{N}{C}\right]$

Such that ${\overrightarrow{F}}_s$ is the actual force in EMF

Power in a Battery

Power supplied by a battery is defined as

$P=IV=I\epsilon$

EMF in a Battery

The total force acting on a charge in a battery is defined as:

${\overrightarrow{F}}_{net}=q(\overrightarrow{E}+{\overrightarrow{f}}_s)$

EMF in a Circuit

Due to the internal resistivity of a battery, $r$:

$\epsilon =I(R+r)$

Thus,

$I=\frac{\epsilon }{R+r}$

Therefore, can be written as:

$V_{battery}=IR=\epsilon -Ir$

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Circuits

Open Circuits

If external is taken to infinity in a circuit, we can use EMF to find:

$I=0$

Thus

$V_{battery}=\epsilon -0=\epsilon$

I.e. a circuit with infinite external resistance causes current to go to 0 which is just like an open circuit, so the of the battery is itself

Short Circuit

If resistance goes to 0, current is described as

$I_{max}=\frac{\epsilon }{r}$

Because internal resistance is small, it is equivalent to the EMF, thus

$V=\epsilon -\epsilon =0$

This is a shorted circuit, so the power is given as

$P_{max}=\frac{{ϵ}^2}{r}$