Study Notes on Resistors in Series and Parallel

Most circuits have more than one component called a resistor that limits the flow of charge in the circuit.
A measure of this limit on charge flow is called resistance.
The simplest combinations of resistors are the series and parallel connections illustrated in Figure 21.2.
The total resistance of a combination of resistors depends on both their individual values and how they are connected.

Resistors are in series whenever the flow of charge (the current) must flow through devices sequentially.

Example: If current flows through a person holding a screwdriver into the Earth:
- Resistance of the screwdriver's shaft
- Resistance of its handle
- Person’s body resistance
- Resistance of her shoes

It is reasonable to sum the individual resistances because the current must pass through each resistor sequentially.
Total resistance (Rtotal) in series can be calculated as: Rs = R1 + R2 + R_3 + …

According to Ohm’s law, the voltage drop (V) across a resistor is calculated using the equation:
V = I imes R
where:
- I = current in amps (A)
- R = resistance in ohms (Ω)
The voltage drop across each resistor in a series connection can be expressed as follows:
V1 = I imes R1
V2 = I imes R2
V3 = I imes R3
The sum of these voltages equals the total voltage output of the source:
V = V1 + V2 + V_3

The derivations for the equations of series resistance are based on energy and charge conservation laws which state:
- Total charge and energy are constant in any process.

Resistors are in parallel when each resistor is connected directly to the voltage source without any intervening resistors affecting the flow, resulting in negligible resistance among the connecting wires.
Each resistor thus has the full voltage of the source applied to it.

The same voltage applies across each resistor in parallel, but the current divides among the resistors:
I = I1 + I2 + I_3

The formula for total or equivalent resistance in a parallel circuit can be expressed as:
rac{1}{Rp} = rac{1}{R1} + rac{1}{R2} + rac{1}{R3} + …
This implies that total resistance (R_total) in parallel is always less than the smallest resistance in the group, reflecting that more current flows out of the source than would with individual resistors due to the total resistance being lower.

Given resistances:
- Voltage output of the battery: 12.0 V
- Resistors:
- $R_1 = 1.00 \, \Omega$
- $R_2 = 6.00 \, \Omega$
- $R_3 = 13.0 \, \Omega$
Total resistance:
Rs = R1 + R2 + R3 = 1.00 + 6.00 + 13.0 = 20.0 \, \Omega
Current in the circuit:
I = \frac{V}{R_s} = \frac{12.0 \, V}{20.0 \, \Omega} = 0.600 \, A
Voltage drop across each resistor:
- For $R1$: V1 = I \times R_1 = (0.600 \, A)(1.00 \, \Omega) = 0.600 \, V
- For $R2$: V2 = I \times R_2 = (0.600 \, A)(6.00 \, \Omega) = 3.60 \, V
- For $R3$: V3 = I \times R_3 = (0.600 \, A)(13.0 \, \Omega) = 7.80 \, V
Total voltage drop check:
V{total} = V1 + V2 + V3 = 0.600 + 3.60 + 7.80 = 12.0 \, V

Power dissipated by each resistor can be calculated using:
P = I^2 imes R
Power for each resistor:
- For $R1$: P1 = I^2 R_1 = (0.600 A)^2 (1.00 \, \Omega) = 0.360 \, W
- For $R2$: P2 = (0.600 A)^2 (6.00 \, \Omega) = 2.16 \, W
- For $R3$: P3 = (0.600 A)^2 (13.0 \, \Omega) = 4.68 \, W
Total power:
P{total} = P1 + P2 + P3 = 0.360 + 2.16 + 4.68 = 7.20 \, W

Resistance in wires, when significant, can lead to lower current and power delivered to devices, impacting performance.
Observations in real-world scenarios:
- Lights dim when a large appliance is switched on due to increased current causing voltage drops in connecting wires.

Draw a clear circuit diagram labeling all resistors and voltage sources.
Identify what needs to be determined in the problem (list unknowns).
Determine whether resistors are in series, parallel, or a combination.
Apply appropriate equations for series and parallel resistors to solve for unknowns.
Check for consistency and reasonableness of answers.