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| 184_notes:examples:week8_resistors_series [2017/10/10 16:03] – [Example: Resistors in Series] tallpaul | 184_notes:examples:week8_resistors_series [2021/07/05 21:45] – schram45 | ||
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| =====Example: | =====Example: | ||
| Suppose you have the following circuit. Resistors are labeled 1 through 3 for convenience of reference. You know that the circuit contains a 12-Volt battery, and $R_1=10 \Omega$, $\Delta V_3=6 \text{ V}$, and the power dissipated through Resistor 1 is $P_1 = 0.1 \text{ W}$. What is the resistance of and power dissipated through Resistor 2? | Suppose you have the following circuit. Resistors are labeled 1 through 3 for convenience of reference. You know that the circuit contains a 12-Volt battery, and $R_1=10 \Omega$, $\Delta V_3=6 \text{ V}$, and the power dissipated through Resistor 1 is $P_1 = 0.1 \text{ W}$. What is the resistance of and power dissipated through Resistor 2? | ||
| - | {{ 184_notes: | + | [{{ 184_notes: |
| ===Facts=== | ===Facts=== | ||
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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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| * We represent the situation with diagram given above. | * We represent the situation with diagram given above. | ||
| ====Solution==== | ====Solution==== | ||
| - | Shortly, we will constrain our calculations to just Resistors 1 and 2. We don't have any information on Resistor 2, so our approach will be to find the equivalent resistance of 1 and 2, and then focus on just Resistor 2 using equation (3). The first steps in our approach will be to find the current and potential difference across these two resistors. //Note, this is not the only approach that would work! Another way would be to find individual potential difference across each resistor, and then focusing on Resistor 2 from there. See if you can think of yet another method...// | + | Shortly, we will constrain our calculations to just Resistors 1 and 2. We don't have any information on Resistor 2, so our approach will be to find the equivalent resistance of 1 and 2, and then focus on just Resistor 2 using equation (3). The first steps in our approach will be to find the current and potential difference across these two resistors. //Note, this is not the only approach that would work! Another way would be to find individual potential difference across each resistor, and then focusing on Resistor 2 from there.// (See if you can think of yet another method...) |
| We can use the Loop Rule -- equation (4) -- to find the potential difference across these two resistors. The potential difference across the battery has opposite sign as the differences across the resistors, if we consider the circuit as a loop of individual differences. We write: | We can use the Loop Rule -- equation (4) -- to find the potential difference across these two resistors. The potential difference across the battery has opposite sign as the differences across the resistors, if we consider the circuit as a loop of individual differences. We write: | ||
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| Since we know $\Delta V_{bat}$ and $\Delta V_3$, we can plug in and solve: $\Delta V_1 + \Delta V_2 = 6 \text{ V}$. | Since we know $\Delta V_{bat}$ and $\Delta V_3$, we can plug in and solve: $\Delta V_1 + \Delta V_2 = 6 \text{ V}$. | ||
| - | We can find current through the circuit using equations | + | Plugging equation |
| Since we know $P_1$ and $R_1$, we can plug in and solve for $I_1=\sqrt{P_1/ | Since we know $P_1$ and $R_1$, we can plug in and solve for $I_1=\sqrt{P_1/ | ||
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| Now, equation (3) tells us $R_{eq}=R_1+R_2$, | Now, equation (3) tells us $R_{eq}=R_1+R_2$, | ||
| 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: | ||
| - | $$P_2=I^2R_2= 0.6 \text{ W}$$ | + | $$P_2=I^2R_2= 0.5 \text{ W}$$ |
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| + | 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*} | ||