Lecture 4: Notes: Electric Circuits - Current, Voltage, Capacitors, Resistors, and Safety

Measuring Current and Voltage

• Measuring current requires interrupting the circuit to insert an ammeter in series.
• Measuring voltage can be done without interrupting the circuit by connecting a voltmeter in parallel, provided the meter has very high resistance.

Series and Parallel Circuits

• In a series circuit, the current is the same everywhere.
• In a parallel circuit, the voltage is the same everywhere.
• To measure current, an ammeter is placed in series with the circuit.
• To measure voltage, a voltmeter is placed in parallel with the circuit.

Ammeter Usage

• To measure the current at a specific point, the ammeter must be inserted at that point, requiring the circuit to be interrupted.
• The current must flow into and out of the ammeter for an accurate reading.

Voltmeter Usage

• To measure the voltage between two points, connect the voltmeter in parallel between those points.
• Interrupting the circuit is not required when measuring voltage.

Voltage Measurement Considerations

• Asking about the voltage at a single point is ambiguous; voltage is always a difference between two points.
• A reference point (e.g., the negative terminal of a battery) is needed to define what zero voltage is.
• Analogy: Asking about the height of a building requires specifying the reference point (ground level vs. sea level).

Circuit Analysis with Meters

• Correct meter usage is crucial for accurate measurements.
• Ammeters must be in series, and voltmeters must be in parallel.

Capacitors in Parallel

• Consider a capacitor with charge $q$ and voltage $v$.
• If this capacitor is split into two parallel capacitors, each has the same voltage $v$.
• The total charge $q$ is divided into two charges, $q<em>1$ and $q</em>2$, such that $q = q<em>1 + q</em>2$.
• Capacitance is defined as $C = \frac{q}{v}$.
• For the split capacitors, $C<em>1 = \frac{q</em>1}{v}$ and $C<em>2 = \frac{q</em>2}{v}$.
• The effective capacitance $C$ of two parallel capacitors is the sum of their individual capacitances: $C = C<em>1 + C</em>2$.

Capacitors in Series

• In a series configuration, two capacitors share the same charge $q$.

• The voltage across the series combination is divided into two steps, $v<em>1$ and $v</em>2$, such that $v = v<em>1 + v</em>2$.

• Since $v = \frac{q}{C}$, we have $v<em>1 = \frac{q}{C</em>1}$ and $v<em>2 = \frac{q}{C</em>2}$.

• Therefore, $\frac{q}{C} = \frac{q}{C<em>1} + \frac{q}{C</em>2}$.

• The effective capacitance $C$ of two capacitors in series is given by:

  $\frac{1}{C} = \frac{1}{C<em>1} + \frac{1}{C</em>2}$

• Generalization for multiple capacitors in series:

  $\frac{1}{C} = \frac{1}{C<em>1} + \frac{1}{C</em>2} + \frac{1}{C_3} + …$

Resistors in Series

• The voltage across series resistors is divided into steps: $v = v<em>1 + v</em>2$.
• Ohm's law: $V = IR$.
• Therefore, $IR = I<em>1R</em>1 + I<em>2R</em>2$.
• In a series circuit, the current is the same: $I = I<em>1 = I</em>2$.
• The effective resistance $R$ of two series resistors is the sum of their individual resistances: $R = R<em>1 + R</em>2$.

Resistors in Parallel

• In parallel circuits, currents add up, and voltages are the same.

• $I = I<em>1 + I</em>2$.

• Using Ohm's law, $I = \frac{V}{R}$, $I<em>1 = \frac{V</em>1}{R<em>1}$, and $I</em>2 = \frac{V<em>2}{R</em>2}$.

• Therefore, $\frac{V}{R} = \frac{V<em>1}{R</em>1} + \frac{V<em>2}{R</em>2}$.

• In a parallel circuit, $V = V<em>1 = V</em>2$.

• The effective resistance $R$ of two parallel resistors is given by:

  $\frac{1}{R} = \frac{1}{R<em>1} + \frac{1}{R</em>2}$

• Generalization for multiple resistors in parallel:

  $\frac{1}{R} = \frac{1}{R<em>1} + \frac{1}{R</em>2} + \frac{1}{R_3} + …$

Example: Capacitance Calculation

• A circuit with three capacitors: 10 microfarads, 2 microfarads, and 3 microfarads.

• First, combine the parallel capacitors: 2 microfarads + 3 microfarads = 5 microfarads.

• Now, combine the series capacitors (10 microfarads and 5 microfarads) by adding reciprocals:

  $\frac{1}{C} = \frac{1}{10} + \frac{1}{5} = \frac{3}{10}$

• Therefore, the equivalent capacitance $C = \frac{10}{3} = 3.33$ microfarads.

Ammeters and Voltmeters: Impact on Circuits

• Ammeters should have very low resistance to avoid affecting the total current in the circuit.
• If an ammeter's resistance is too high, it will reduce the current being measured.
• Voltmeters should have very high resistance to avoid current flowing through them, which would alter the voltage being measured.

Electrical Safety

• Muscles are triggered by electrical stimulation.
• Alessandro Volta's experiments demonstrated muscle contractions induced by electricity.
• Electric shock can cause muscles to twitch and, at higher currents, can prevent a person from releasing an object.
• High currents can cause heart fibrillation (irregular heartbeat), which can be dangerous.
• Very high currents can cause the heart to seize or lead to overheating.
• The heart is particularly vulnerable; fibrillation is a significant danger.
• The current must pass through the body to cause harm.
• The path the current takes through the body is critical.
• The idea of dropping electrical appliances into bathtubs is usually not fatal, as the water provides a better conductive path than a person's body, so current will pass through the water, not the body.