Series and Parallel Resistors Study Notes

Definition: Resistors are said to be in series when they are connected end-to-end, such that the same current flows through each resistor.

Current Rule: In a series circuit, the current (I) through each resistor is the same. This is due to the single pathway for current flow.
Equation: $I{total} = I{R1} = I{R2} = I{R3}$

Voltage Rule: The total voltage across the circuit is equal to the sum of the voltages across each resistor.
Equation: $V{total} = V{R1} + V{R2} + V{R3}$

Total Resistance: The total resistance in a series circuit is the sum of the individual resistances.
Equation: $R{total} = R{1} + R{2} + R{3}$

Power Calculation: The power (P) used by each resistor can be calculated using the formula: $P = I^2 imes R$

Definition: Resistors are in parallel when they are connected such that both terminals of each resistor are connected to the same two points, creating multiple pathways for current flow.

Current Rule: The total current entering the parallel circuit is equal to the sum of the currents through each resistor.
Equation: $I{total} = I{R1} + I{R2} + I{R3}$

Voltage Rule: The voltage across each resistor in a parallel circuit is the same and is equal to the voltage of the source.
Equation: $V{R1} = V{R2} = V{R3} = V{total}$

Total Resistance: The total resistance of a parallel circuit can be calculated using the formula:
Equation: $\frac{1}{R{total}} = \frac{1}{R{1}} + \frac{1}{R{2}} + \frac{1}{R{3}}$

Power Calculation: The power (P) consumed by each parallel resistor can be determined using the formula: $P = \frac{V^2}{R}$

Voltage Across Resistors: When a resistor is added to a series circuit, the voltage across each resistor will divide because the voltage from the battery is distributed between the resistors. The total voltage drop across all of the resistors will always be equal to the voltage of the battery.
Total Current: When a resistor is added to a series circuit, the total current through the circuit decreases because the total resistance increases.
Total Resistance Impact: When adding a resistor, the total resistance of the circuit increases, and this leads to a decrease in the current.
Brightness of Bulbs: When a bulb is added to a circuit in series, the brightness of the other bulbs decreases because the total current of the circuit is divided between the bulbs.

Voltage Across Resistors: When a resistor is added to a parallel circuit, the voltage across each of the resistors will remain the same because each resistor is connected directly to the battery.
Total Current Increase: When a resistor is added to a parallel circuit, the total current through the circuit increases because there are more pathways for the current.
Total Resistance Impact: When a resistor is added to a parallel circuit, the total resistance of the circuit decreases, leading to an increase in current.
Brightness of Bulbs: When a bulb is added to a circuit in parallel, the brightness of the other bulbs remains unchanged because the voltage across each bulb remains the same.

Current (I): State the current rule for resistors in series: Current is constant throughout the series circuit.
Voltage (V): State the voltage rule for resistors in series: The total voltage is distributed among the resistors, equaling the battery voltage.
Resistance (R): Total resistance is calculated by summing all individual resistances.
Power (Bulb Brightness): Power output of the individual bulbs in series can be calculated via $P = I^2 imes R$ .

Given Information: The circuit contains three resistors connected in series. The voltage of the source $VB = 12 ext{ V}$ . The resistances are: $R1 = 5 ext{ } ext{Ω}, R2 = 9 ext{ } ext{Ω}, R3 = 7 ext{ } ext{Ω}$ .
Total Resistance Calculation:
- $R{total} = R1 + R2 + R3$
- $R_{total} = 5 + 9 + 7 = 21 ext{ Ω}$
Total Current Calculation: Using Ohm's Law, $I = \frac{V}{R}$ ,
- $I = \frac{12 ext{ V}}{21 ext{ Ω}} = 0.571 ext{ A}$
Current Through Each Resistor:
- $I{R1} = I{R2} = I_{R3} = 0.571 ext{ A}$
Voltage Drops:
- $V{R1} = I imes R1 = 0.571 imes 5 = 2.857 ext{ V}$
- $V{R2} = I imes R2 = 0.571 imes 9 = 5.143 ext{ V}$
- $V{R3} = I imes R3 = 0.571 imes 7 = 4.0 ext{ V}$
Power Ratings:
- $P{R1} = I^2 imes R1 = (0.571)^2 imes 5 ext{ Ω} = 1.63 ext{ W}$
- $P_{R2} = (0.571)^2 imes 9 = 2.95 ext{ W}$
- $P_{R3} = (0.571)^2 imes 7 = 2.04 ext{ W}$

Circuit 1: Resistors R1, R2, R3 with equal resistance; Voltage source $V_S = 18 ext{ V}$ .
- Given $I_2 = 0.75 ext{ A}$ .
- Determine missing voltage $V1$ , $V2$ , and current values as applicable.
Circuit 2: Resistors R1, R2, R3 with varying resistance having voltage source $V_S = 7 ext{ V}$ .
- Given $V1 = 2.0 ext{ V}, V3 = 3.5 ext{ V}, I_C = 0.5 ext{ A}$ .
- Complete missing current and voltage values as applicable.

Total Resistance: When a switch across points A and B is opened:
- Total resistance $R_T$ increases because it eliminates a parallel path.
Total Current: Total current $I_T$ through the circuit will decrease due to increased resistance.
Voltage Drop Across Bulbs: Voltage drops across Bulbs 2 and 3 will increase as the current divides among fewer pathways.
Current Through Bulbs: Current through Bulbs 2 and 3 will decrease due to decreased total current.
Bulb Brightness: Brightness of Bulbs 2 and 3 will decrease due to decreased current.

Complete the circuit with voltage source $V_S$ and bulbs in series.
Use volt meters to measure voltage gain across the source and voltage drops across each bulb. Measure: $V1, V2, V3, VS$ .

Understand Bulb Wiring: Using provided data on current, voltage drop, and resistance across three connected bulbs, analyze circuit behavior when a bulb is removed.
Voltage of Battery: Deduce the battery voltage based on cumulative voltage drops across the bulbs.
Current of Battery: Analyze current through the battery, explaining steady conditions under discussed scenarios.
Equivalent Resistance Calculation: Find equivalent resistance when a bulb is removed, explaining variations in current and voltage drop as a result.
Impact of Bulb Removal: Determine the effect on current, voltage drop, and remaining bulb brightness with one bulb removed.

Understanding the principles of series and parallel circuits is crucial for predicting how additional components influence voltage distribution, current flow, and overall circuit behavior. These concepts are foundational for studying more complex electrical systems.