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| 184_notes:examples:week8_resistors_series [2019/03/20 17:23] – [Solution] hallstein | 184_notes:examples:week8_resistors_series [2021/07/05 21:45] (current) – [Solution] schram45 | ||
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| ===Approximations & Assumptions=== | ===Approximations & Assumptions=== | ||
| - | * The wire has very very small resistance when compared to the other resistors in the circuit. | + | * The wire has very very small resistance when compared to the other resistors in the circuit: This allows there to be no energy loss across the wires and no potential difference across them either simplifying down the model. |
| - | * The circuit is in a steady state. | + | * The circuit is in a steady state: It takes a finite amount of time for a circuit to reach steady state and set up a charge gradient. Making this assumption means the current is not changing with time in any branch of the circuit. |
| - | * Approximating the battery as a mechanical battery. | + | * Approximating the battery as a mechanical battery: This means the battery will supply a steady power source to the circuit to keep it in steady state. |
| - | * The resistors in the circuit are made of Ohmic materials. | + | * The resistors in the circuit are made of Ohmic materials: Ohmic materials have a linear relationship between voltage and current, this allows us to use ohms law. |
| ===Representations=== | ===Representations=== | ||
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| The power dissipated across Resistor 2 can be found using the same rewriting of equation (2) as above: | The power dissipated across Resistor 2 can be found using the same rewriting of equation (2) as above: | ||
| $$P_2=I^2R_2= 0.5 \text{ W}$$ | $$P_2=I^2R_2= 0.5 \text{ W}$$ | ||
| + | |||
| + | One way in which we can evaluate our solution in this problem is by seeing if the power generated by the battery is equal to the power dissipated through the resistors. You will find that the 1.2 Watts of power generated by the battery is completely dissipated through the resistors. | ||
| + | \begin{align*} | ||
| + | P_{gen} &= P_{dis} \\ | ||
| + | I\Delta V_{bat} &= P_1 + P_2 + P_3 \\ | ||
| + | I\Delta V_{bat} &= P_1 + P_2 + I\Delta V_3 \\ | ||
| + | 0.1\left(12\right) &= 0.1 + 0.5 + 0.1\left(6\right) \\ | ||
| + | 1.2 &= 1.2 | ||
| + | \end{align*} | ||