184_notes:examples:week7_ohms_law

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184_notes:examples:week7_ohms_law [2017/10/04 18:27] – [Solution] tallpaul184_notes:examples:week7_ohms_law [2018/06/19 14:54] (current) – curdemma
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 +[[184_notes:resistivity|Return to resistors and conductivity]]
 +
 =====Example: Application of Ohm's Law===== =====Example: Application of Ohm's Law=====
 Suppose you have a simple circuit that contains only a 9-Volt battery and a resistor of $120 \Omega$. What is the current in the wire? Suppose you have a simple circuit that contains only a 9-Volt battery and a resistor of $120 \Omega$. What is the current in the wire?
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 ===Approximations & Assumptions=== ===Approximations & Assumptions===
-  * The wire has no (negligible) resistance.+  * The wire has very very small resistance when compared to the 120 $\Omega$ resistor.
   * The circuit is in a steady state.   * The circuit is in a steady state.
   * Approximating the battery as a mechanical battery.   * Approximating the battery as a mechanical battery.
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 ===Representations=== ===Representations===
   * We represent [[184_notes:resistivity#Resistance|Ohm's Law]] as $\Delta V = IR$.   * We represent [[184_notes:resistivity#Resistance|Ohm's Law]] as $\Delta V = IR$.
-  * We represent the situation with following picture.+  * We represent the situation with following circuit diagram.
  
-{{ 184_notes:7_ohm.png?300 |Circuit with Resistor}}+[{{ 184_notes:7_ohm.png?300 |Circuit with Resistor}}]
  
 ====Solution==== ====Solution====
-We have assumed that the battery and the wire contribute negligible resistance to the circuit. So the resistance of the circuit is simply the resistance of the resistor: $R = 120\Omega$. These assumptions also lead us to conclude that the voltage difference across the circuit (from the positive end of the battery to the negative end) is the entire 9 Volts, so $\Delta V = 9 \text{ V}$. We can now use Ohm's Law to find the current through the circuit: $$I = \frac{\Delta V}{R} = 75 \text{ mA}$$+We have assumed that the battery and the wire contribute negligible resistance to the circuit. So the resistance of the circuit is simply the resistance of the resistor: $R = 120\Omega$. These assumptions also lead us to conclude that the voltage that is provided by the battery is used up only in the resistor (and not on the wires). So the potential difference across the resistor is the entire 9 Volts, so $\Delta V = 9 \text{ V}$. We can now use Ohm's Law to find the current through the circuit: $$I = \frac{\Delta V}{R} = 75 \text{ mA}$$
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  • Last modified: 2017/10/04 18:27
  • by tallpaul