What are the Equilibrium voltages of the capacitor and the resistor when the function generator voltage is +v0? Is this what you observe on your oscilloscope sketches? what are the equilibrium voltages of the capacitor and the resistor when the function generator voltage is ± you observe on your oscilloscope sketches? ? Is this what An RC circuit consists of a resistor connected to a capacitor. There may also be a fixed power supply, depending on whether the capacitor is being charged or discharged.1 In this lab, rather than using a power supply to charge and discharge the capacitor, we will be using an input of a square wave voltage, which oscillates between some voltage Vo and -Vo (which depends on on the amplitude you set). This will repeatedly charge the capacitor one way, then the other, with currents passing through the resistor each time We will denote the input voltage from the function generator as Vfc(t), the voltage across the capacitor as Vc(t), and the voltage across the resistor as VR(t) We treat "positive" as from the high side of the function generator to the low side, which implies (by Kirchhoff's Voltage Law) VFG(t) VR(t) Vc(t) We also know that VR (t)I(t)R (Ohm's law) and Vc(t)Q(t)/C (definition of capacitance) Although the details of amplitudes are slightly different than a purely discharging capacitor, one still observes an exponential decay to the equilibrium voltage, with the same time constant: We will make measurements, for both the capacitor and the resistor, of the difference between the voltage across the component and the equilibrium voltage across the component. That is to say, if the capacitor approaches an equilibrium voltage of Vc,eq, then we will measure Vc(t) - Vc,eg (and similarly for the resistor) Call this quantity Vaiff(t). Then, for both the resistor and the capacitor, we expect the relationship (with some initial voltage V): diff There is a practical issue with this from our perspective, however: this is a nonlinear relationship, so our plotting tool cannot infer T from a slope! Therefore, we apply a trick: we make a log-linear plot. By taking a logarithm of both sides of (2), we can reduce this to the equation This means that if we plot In(Viff) vs. t, we will get a linear plot (with intercept), and can infer T from the slope. Problem solved! We were unable to transcribe this image what are the equilibrium voltages of the capacitor and the resistor when the function generator voltage is ± you observe on your oscilloscope sketches? ? Is this what An RC circuit consists of a resistor connected to a capacitor. There may also be a fixed power supply, depending on whether the capacitor is being charged or discharged.1 In this lab, rather than using a power supply to charge and discharge the capacitor, we will be using an input of a square wave voltage, which oscillates between some voltage Vo and -Vo (which depends on on the amplitude you set). This will repeatedly charge the capacitor one way, then the other, with currents passing through the resistor each time We will denote the input voltage from the function generator as Vfc(t), the voltage across the capacitor as Vc(t), and the voltage across the resistor as VR(t) We treat "positive" as from the high side of the function generator to the low side, which implies (by Kirchhoff's Voltage Law) VFG(t) VR(t) Vc(t) We also know that VR (t)I(t)R (Ohm's law) and Vc(t)Q(t)/C (definition of capacitance) Although the details of amplitudes are slightly different than a purely discharging capacitor, one still observes an exponential decay to the equilibrium voltage, with the same time constant: We will make measurements, for both the capacitor and the resistor, of the difference between the voltage across the component and the equilibrium voltage across the component. That is to say, if the capacitor approaches an equilibrium voltage of Vc,eq, then we will measure Vc(t) - Vc,eg (and similarly for the resistor) Call this quantity Vaiff(t). Then, for both the resistor and the capacitor, we expect the relationship (with some initial voltage V): diff There is a practical issue with this from our perspective, however: this is a nonlinear relationship, so our plotting tool cannot infer T from a slope! Therefore, we apply a trick: we make a log-linear plot. By taking a logarithm of both sides of (2), we can reduce this to the equation This means that if we plot In(Viff) vs. t, we will get a linear plot (with intercept), and can infer T from the slope. Problem solved!