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--- 184_notes:examples:week7_ohms_law [2017/10/04 18:20] – [Example: Application of Ohm's Law] tallpaul
+++ 184_notes:examples:week7_ohms_law [2018/06/19 14:54] (current) – curdemma
@@ Line 1: / Line 1: @@
+[[184_notes:resistivity|Return to resistors and conductivity]]
 =====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?
@@ Line 10: / Line 12: @@
 ===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.
   * Approximating the battery as a mechanical battery.
@@ Line 17: / Line 19: @@
 ===Representations===
   * 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====
-Let's start with node  $A$ . Incoming current is  $I_1$ , and outgoing current is  $I_2$ . How do we decide if  $I_{A\rightarrow B}$  is incoming or outgoing? We need to bring it back to the Node Rule: $I_{in}=I_{out}$. Since  $I_1=8 \text{ A}$  and  $I_2=3 \text{ A}$ , we need  $I_{A\rightarrow B}$  to be outgoing to balance. To satisfy the Node Rule, we set
+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: $$I = \frac{\Delta V}{R} = 75 \text{ mA}$$
- $I_{A\rightarrow B} = I_{out}-I_2 = I_{in}-I_2 = I_1-I_2 = 5 \text{ A}$
-We do a similar analysis for node  $B$ . Incoming current is  $I_{A\rightarrow B}$ , and outgoing current is  $I_3$ . Since  $I_{A\rightarrow B}=5 \text{ A}$  and  $I_3=4 \text{ A}$ , we need  $I_{B\rightarrow D}$  to be outgoing to balance. To satisfy the Node Rule, we set
- $I_{B\rightarrow D} = I_{out}-I_3 = I_{in}-I_3 = I_{A\rightarrow B}-I_3 = 1 \text{ A}$
-For node  $C$ , incoming current is  $I_2$  and  $I_3$ . There is no outgoing current defined yet!  $I_{C\rightarrow D}$  must be outgoing to balance. To satisfy the Node Rule, we set
-$$I_{C\rightarrow D} = I_{out} = I_{in} = I_2+I_3 = 7 \text{ A}$$
-Lastly, we look at node $D$. Incoming current is  $I_{B\rightarrow D}$  and  $I_{C\rightarrow D}$ . Since there is no outgoing current defined yet,  $I_{D\rightarrow battery}$  must be outgoing to balance. To satisfy the Node Rule, we set
-$$I_{D\rightarrow battery} = I_{out} = I_{in} = I_{B\rightarrow D}+I_{B\rightarrow D} = 8 \text{ A}$$
-Notice that  $I_{D\rightarrow battery}=I_1$ . This will always be the case for currents going in and out of the battery (approximating a few things that are usually safe to approximate, such as a steady current). In fact, we could have treated the battery as another node in this example. Notice also that if you incorrectly reason about the direction of a current (incoming or outgoing), the calculation will give a negative number for the current. The Node Rule is self-correcting. A final diagram with directions is shown below.
-{{ 184_notes:6_nodes_with_arrows.png?300 |Circuit with Nodes}}