Power Dissipation In a Resistor

Purpose

A resistor that is getting warm is leaking power. This note states the power relations and Kirchhoff’s laws, works one divider example end to end, and models the iPhone standby draw from lecture 2 as a single resistor.

Power dissipation

$P=VI=I^{2}R=V^2/R$

Kirchhoff’s laws

Voltage (KVL): The voltages along any closed path around a circuit sum to zero.

Current (KCL): The current arriving at any node equals the current leaving that node.

Worked example

A 120 V source drives two resistors in series:

      R_1 = 1 Ohm
  +--/\/\/|--+
  |          |
  _          |
  + 120 V    |
  -          |
  |          |
  +--\/\/\|--+
      R_2 = 10 Ohm

By Ohm’s law, $V_1=I{R}_1$ and $V_2=I{R}_2$. The voltage divider gives the voltage across $R_2$:

$V_2=V\cdot \frac{R_2}{R_1+R_2}=120\cdot \frac{10}{11}=109\text{V}$

$P_2=\frac{V_2^2}{R_2}=\frac{109^2}{10}=1188\text{W}$

The current follows from the voltage left across $R_1$:

$I=\frac{V-V_2}{R_1}=\frac{120-109}{1}=11\text{A}$

$P_1=I^2{R}_1=11^2\times 1=121\text{W}$

The divider formula assumes no load

$V_2=V\cdot \frac{R_2}{R_1+R_2}$ holds only while nothing draws current from the node between $R_1$ and $R_2$. A load attached there sits in parallel with $R_2$, which lowers the effective resistance and pulls the output voltage down. Recompute with $R_2$ replaced by $R_2\parallel R_{load}$.

Ideal sources

A voltage source produces a constant voltage regardless of the current drawn, which is an idealistic assumption. A current source delivers constant current to a circuit, with the voltage across it determined by the load resistance (impedance).

Modeling the iPhone as a resistor

In lecture 2 we estimated that an iPhone in standby draws 7.55 mA from its 3.8 V lithium battery. Model the phone’s power consumption as a single resistor across the battery:

 
ground
  ^
  |     battery
  |       |
  |       |
  +----- |-  +|-----+
  |                 |
  |                 |
  +-----|/\/\/|-----+
           |
           |
 phone power consumption
 

The equivalent resistance is $R=V/I=3.8\text{V}/7.55\text{mA}\approx 503\Omega$.

Related

• Resistance
• Thevenin’s Theorem
• Electric Circuit Analysis
• Electronic Components