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PHYS 102

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Apr 3, 2024

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PHYS 102 Discussion 8: Group activity Kalyssa Smith, Juno Jang, Isis Johnson TOTAL POINTS 4.65 / 5 QUESTION 1 1 Preparation: Loops, Junctions 1 / 1 ✓ + 0.25 pts Loop explanation ✓ + 0.25 pts Junction explanation ✓ + 0.25 pts Example circuits ✓ + 0.25 pts Explain how voltage/current across resistor come about QUESTION 2 2 Preparation: RC 0.75 / 1 ✓ + 0.25 pts Example circuit + 0.25 pts Q, I vs t ✓ + 0.25 pts how fast will C charge? ✓ + 0.25 pts Plots After closing the switch, we use the loop rule: ε -Q/C-IR=0. Initially, the charge is 0, and the current is maximal. Then, the capacitor will be charged up and the current decreases at the same time: The charge on the capacitor is described as Q(t)=Q_∞ (1-e^((-t)⁄RC) ), where Q_∞= ε C. We can see that the time the charging takes is determined by the values for R and C. To compute the time it takes to charge to 99% of Q_∞, we have to solve the equation Q(t)=0.99Q_∞=Q_∞ (1-e^((-t)⁄RC) ), which can be solved to t=-RC ln ⁡ (0.01). QUESTION 3 3 Team A: Currents 1 / 1 ✓ + 0.25 pts Approach ✓ + 0.25 pts Big ideas, formulas ✓ + 0.5 pts Solution and verification We need to use the ideas of resistors in series and parallel, Ohm’s law, and Kirchhoff’s rules. To find the current through R_3, we will take the following steps: First, we know from the loop rule that ε -I_2 R_2-I_1 R_1=0, and that I_2 R_2=I_3 R_3. We also know that I_1=I_2+I_3 from the junction rule. Combining these equations, we can solve for I_3, because we have three equations for three unknowns (the currents). The result is I_3=( ε R_2)/(R_1 R_2+R_3 R_2+R_1 R_3 ) We see that we probably want to have a large value for R_2 to get a large current. In addition, picking a small R_1 will make the denominator smaller, and

increase the current as well. We pick ± R_1=1 Ω and R_2=100 Ω . With these numbers, I_3=0.45 A, which meets our objective. QUESTION 4 4 Team B: Batteries, switches 1 / 1 ✓ + 0.25 pts Approach ✓ + 0.25 pts Big ideas, formulas ✓ + 0.5 pts Solution, verification Because of R_A, the battery will not provide voltage to generate the required current through R_B. The key is to discharge the capacitor instead, without it being connected to R_A. We pick R_A=100 Ω , and C=10 μ F . (These values don’t matter for this activity, only for the next one.) The first step is to close S1 and leave S2 open. After a long time, the capacitor is charged to 5 V. After that, we open S1, and then close S2. This RC circuit will lead to the capacitor being discharged through the resistor. We know that the current through the resistor will be given by I(t)=I_0 e^(-t/RC) And we know that I_0=Q_0/RC=V/R That means, right after closing the switch, the current is I_0=(5 V)/(10 Ω )=0.5 A. QUESTION 5 5 Team C: discharge 0.9 / 1 + 0.1 pts New circuit ✓ + 0.2 pts Approach ✓ + 0.2 pts Big ideas, formulas ✓ + 0.5 pts Solution, verification + 0 pts missing This is a RC circuit as seen before. The current through R_3 immediately after connecting the capacitor is I_3,0=0.45 A, as computed in activity 3. After that, the current will decrease exponentially, as I_3 (t)=I_3,0 e^(-t/RC) The resistance we have to use here is the equivalent resistance of the circuit, because that is the resistance that determines how fast the capacitor is discharged. We can evaluate it as R=R_1+1/(1/R_2 +1/R_3 )=10.1 Ω And RC=10.1 ⋅ 10 × 10^(-6) s=0.101 ms

To find the time it takes to reach 1 nA of current through R_3 we have to solve 1 × 10^(-9) A= I_3,0 e^(-t/RC) Which gives us: ln ⁡ ((1 × 10^(-9))/0.45)=-t/RC t=-RC ln ⁡ ((1 × 10^(-9))/0.45) =2.01 ms Page 3

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1 Preparation: Loops, Junctions 1 / 1 ✓ + 0.25 pts Loop explanation ✓ + 0.25 pts Junction explanation ✓ + 0.25 pts Example circuits ✓ + 0.25 pts Explain how voltage/current across resistor come about Page 5

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2 Preparation: RC 0.75 / 1 ✓ + 0.25 pts Example circuit + 0.25 pts Q, I vs t ✓ + 0.25 pts how fast will C charge? ✓ + 0.25 pts Plots After closing the switch, we use the loop rule: ε -Q/C-IR=0. Initially, the charge is 0, and the current is maximal. Then, the capacitor will be charged up and the current decreases at the same time: The charge on the capacitor is described as Q(t)=Q_∞ (1-e^((-t)⁄RC) ), where Q_∞= ε C. We can see that the time the charging takes is determined by the values for R and C. To compute the time it takes to charge to 99% of Q_∞, we have to solve the equation Q(t)=0.99Q_∞=Q_∞ (1-e^((-t)⁄RC) ), which can be solved to t=-RC ln ⁡ (0.01). Page 7

3 Team A: Currents 1 / 1 ✓ + 0.25 pts Approach ✓ + 0.25 pts Big ideas, formulas ✓ + 0.5 pts Solution and verification We need to use the ideas of resistors in series and parallel, Ohm’s law, and Kirchhoff’s rules. To find the current through R_3, we will take the following steps: First, we know from the loop rule that ε -I_2 R_2-I_1 R_1=0, and that I_2 R_2=I_3 R_3. We also know that I_1=I_2+I_3 from the junction rule. Combining these equations, we can solve for I_3, because we have three equations for three unknowns (the currents). The result is I_3=( ε R_2)/(R_1 R_2+R_3 R_2+R_1 R_3 ) We see that we probably want to have a large value for R_2 to get a large current. In addition, picking a small R_1 will make the denominator smaller, and increase the current as well. We pick ± R_1=1 Ω and R_2=100 Ω . With these numbers, I_3=0.45 A, which meets our objective. Page 9

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4 Team B: Batteries, switches 1 / 1 ✓ + 0.25 pts Approach ✓ + 0.25 pts Big ideas, formulas ✓ + 0.5 pts Solution, verification Because of R_A, the battery will not provide voltage to generate the required current through R_B. The key is to discharge the capacitor instead, without it being connected to R_A. We pick R_A=100 Ω , and C=10 μ F . (These values don’t matter for this activity, only for the next one.) The first step is to close S1 and leave S2 open. After a long time, the capacitor is charged to 5 V. After that, we open S1, and then close S2. This RC circuit will lead to the capacitor being discharged through the resistor. We know that the current through the resistor will be given by I(t)=I_0 e^(-t/RC) And we know that I_0=Q_0/RC=V/R That means, right after closing the switch, the current is I_0=(5 V)/(10 Ω )=0.5 A. Page 12

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5 Team C: discharge 0.9 / 1 + 0.1 pts New circuit ✓ + 0.2 pts Approach ✓ + 0.2 pts Big ideas, formulas ✓ + 0.5 pts Solution, verification + 0 pts missing This is a RC circuit as seen before. The current through R_3 immediately after connecting the capacitor is I_3,0=0.45 A, as computed in activity 3. After that, the current will decrease exponentially, as I_3 (t)=I_3,0 e^(-t/RC) The resistance we have to use here is the equivalent resistance of the circuit, because that is the resistance that determines how fast the capacitor is discharged. We can evaluate it as R=R_1+1/(1/R_2 +1/R_3 )=10.1 Ω And RC=10.1 ⋅ 10 × 10^(-6) s=0.101 ms To find the time it takes to reach 1 nA of current through R_3 we have to solve 1 × 10^(-9) A= I_3,0 e^(-t/RC) Which gives us: ln ⁡ ((1 × 10^(-9))/0.45)=-t/RC t=-RC ln ⁡ ((1 × 10^(-9))/0.45) =2.01 ms Page 14