# 03-01: Voltage Dividers #

## Voltage Divider #

- The voltage divider circuit consists of a voltage source (.$V_S$) and two resistors (.$R_1$ and .$R_2$)
- Example:4. Find expressions for element currents for all elements (except the voltage source) – all steps on page 3
- We label the node connected to the voltage supply as .$u_1 (= V_S)$, since the voltage supply goes between this node and ground.
- Label the remaining node as .$u_\text{mid}$ and the voltages and currents through every element in the circuit with .$V_i$ and .$I_i$ respectively
- Write KCL equations for all nodes with unknown voltage - in this case, this is just .$u_\text{mid}$, since .$u_1 = V_S$.
- The current entering that node is .$I_{R_1}$ and the current leaving it is .$I_{R_2}$
- Since these currents must be equal, .$I_{R_1} = I_{R_2}$

5. Substitute the element currents into our KCL equation. We have $$I_{R_1} = I_{R_2} \Longrightarrow \frac{V_S - u_\text{mid}}{R_1} = \frac{u_\text{mid}}{R_2}$$ 7. Solve the above equation. Rearranging, we find that $$V_S R_2 âˆ’u_\text{mid}R_2 = u_\text{mid}R_1$$ $$ \Longrightarrow u_\text{mid}(R_1 +R_2) = V_SR_2$$ $$ \Longrightarrow u_\text{mid} = \frac{R_2}{R_1 + R_2} V_S = \frac{1}{1+ \frac{R_1}{R_2}} V_S = \alpha V_S$$$$I_{R_1} = \frac{V_S - u_\text{mid}}{R_1}$$

$$I_{R_2} = \frac{u_\text{mid}}{R_2}$$

The reason this circuit is called a “voltage divider” is that

*we can create any output voltage*of .$u_\text{mid} = \alpha V_S$ for any .$\alpha \in [0,1]$ (assuming that all of the resistance values are non-negative) by varying the ratio of the resistor values .$R_1/R_2$. As we will see shortly, varying this ratio is exactly the mechanism we will use to convert the relative position of a userâ€™s touch to a voltage.

.$R_2$, the

resistor in the numerator, is the one next to ground. .$R_1$ is connected to a non-zero voltage node (in this case .$u_1 = V_S$).

## Capacitor Divider #

The capacitor divider is similar, differing in that the numerator is now the component closest to $V_{in}$ rather than closest to ground (as in the voltage divider with resistors) $$V_{out} = \frac{C_1}{C_1 + C_2} V_{in}$$

## Current Divider #

Current $I_X$ in a resistor $R_X$ that is in parallel with a combination of other resistors of total resistance $R_T$ is $$I_X = \frac{R_T}{R_X + R_T}I_T$$

- $I_T$ is the total current entering the combined network of $R_X$ in parallel with $R_T$
- $R_T$: Total resistance of the circuit to the right of resistor $R_X$

# 03-03: Power and Voltage/Current Measurement #

## Physics of Circuits #

- Power refers to the rate of energy change, .$P = \frac{dE}{dt} \text{ [Watts]}$
- .$\dots \Longrightarrow dE = V\ dQ \Longrightarrow \frac{dE}{dt} \equiv P = V \frac{dQ}{dt} = VI$
- .$P_\text{el} = I_\text{el} \cdot V_\text{el} = V^2_\text{el} \cdot R^{-1}_\text{el} = I^2 _\text{el} R _\text{el}$

- $P > 0 \implies$ (positive) power
*dissipated*, negative power*generated*.- By PSC, resistors always dissipate power because current enters the .$+$ terminal
- Voltage sources tend to generate power, since current comes out of the .$+$ terminal (and the product .$P = IV < 0$)

- $P < 0 \implies$ (positive) power
*generated*, negative power*dissipated*. - In an
isolated (circuit) system, the sum of the power (across all components) should equal zero by conservation of energy
- Useful sanity-check

- .$\dots \Longrightarrow dE = V\ dQ \Longrightarrow \frac{dE}{dt} \equiv P = V \frac{dQ}{dt} = VI$
- .$R = \rho \frac{L}{A}$

## Touchscreen #

- Given that the top (red) layer has a resistivity .$\rho$ and a cross-sectional area .$A$, the resistance of the top layer from the touchpoint to the right-hand end is given by .$R_1 = \rho \frac{L_\text{rest}}{A}$, the resistance of the top layer from the left-hand end to the touchpoint is given by .$R_2 = \rho \frac{L_\text{touch}}{A}$
- We can see that .$u_\text{mid}$ can be found because it’s a voltage divider:
$$u_\text{mid} = \frac{\rho \frac{L_\text{touch}}{A}}{\rho \frac{L_\text{rest}}{A} + \rho \frac{L_\text{touch}}{A}}V_S = \frac{L_\text{touch}}{L_\text{touch} + L_\text{rest}}V_S = \frac{L_\text{touch}}{L}V_S$$
- .$L = L_\text{touch} + L_\text{rest}$: The length of the touchable portion of the screen

The relationship we have found between .$u_\text{mid}$ and .$V_S$ is very convenient because .$u_\text{mid}$ is not dependent on any material property such as .$\rho$ and .$A$. This means that the top layer can be built with any material and the relationship between .$u_\text{mid}$ and .$V_S$ is still valid. There are always some non-idealities in the world – by making .$u_\text{mid}$ independent of any material property, we can make the circuit model immune to such non-idealities. We also have the freedom to choose a material for the top layer that is good for display purposes (rather than needing a specific material for the touchscreen to work).

## Measuring a Circuit #

The voltmeter measures voltage across the circuit, while the ammeter needs to be put in-line with the circuit so that the current flows through the ammeter.

**The measurement should not change the energy of the circuit**- It turns out that the most complete and concise way of guaranteeing these measurement tools do not influence the circuit is to state that
**they do not allow any power dissipated through the measurement device.**

- It turns out that the most complete and concise way of guaranteeing these measurement tools do not influence the circuit is to state that

### Voltmeter #

Because our voltmeter is made to measure voltage, we can naturally assume that the .$V \neq 0$; this means that a voltmeter must have .$I = 0$ going into it to ensure .$P = IV = 0$.

.$I=0$ occurs when in open-circuits, where exactly zero current is flowing.

Recall that, for a given voltage, the higher the associated resistance, the lower the current and therefore the lower the dissipated power. That is, .$\lim{R \to \infty} \Longrightarrow I = 0$

$$V_\text{el1} = I_\text{meas} R$$ $$V_? - V_\text{el1} - V_\text{meas} = 0$$ $$\Longrightarrow V_? = V_\text{el1} + V_\text{meas}$$ $$\Longrightarrow V_? = I_\text{meas} R + V_\text{meas}$$ $$\therefore V_? = V_\text{meas} \iff I_\text{meas} = 0$$

$$I_\text{meas} = \frac{V_\text{meas}}{R_\text{meas}}$$ $$\therefore I_\text{meas} = 0 \iff R_\text{meas} \gg V_\text{meas}$$

Voltmeters are added in

parallelto the circuit, otherwise they would stop the current from flowing.

### Ammeter #

The ammeter has the circuitâ€™s current flowing through it. Therefore, to ensure .$P = IV = 0$, the ammeter needs .$V = 0$.

.$V=0$ occurs in short-circuits (ideal wires), where exactly zero potential difference exists

$$I_? = I_R + I_\text{meas}$$ $$\therefore I_? = I_\text{meas} \iff I_R = 0$$

$$I_R = \frac{V_\text{meas}}{R}$$ $$\therefore I_R = 0 \iff V_\text{meas} = 0$$

Ammeters are added in

seriesto the circuit, otherwise they would short-circuit the measured component.