Simple Circuits: Series and Parallel

Simple Circuits

Simple Circuits: Series and Parallel

In a series circuit, current follows a single path.
In a parallel circuit, current follows more than one path.

Essential Questions

What are the characteristics of series and parallel circuits?
How are currents, potential differences, and equivalent resistances in series circuits related?
How are currents, potential differences, and equivalent resistances in parallel circuits related?

Vocabulary

Resistance
Series circuit
Equivalent resistance
Voltage divider
Parallel circuit

River Model for Electric Circuits

A mountain river can be used to model an electric circuit.
Similarities:
- Water flows downhill ↔ Positive charges move from high to low potential.
- One or more streams ↔ One or more current paths.
- Rocks hinder water flow ↔ Resistors.
- Solar energy evaporates water, clouds precipitate ↔ Battery moves charges from low to high potential.

Series Circuits

A series circuit has only one path for the current.
The current is the same through the entire series circuit.
The increase in voltage provided by the energy source (\Delta V_{source}) equals the sum of voltage drops across resistors.
The current through the circuit is constant.
A single resistor (R) can replace the series of resistors, having a resistance equal to the sum of individual resistances; this is the equivalent resistance.
The equivalent resistance of resistors in series equals the sum of the individual resistances of the resistors.
The equivalent resistance is greater than that of any individual resistor.

Equivalent Resistance for Resistors in Series

The equivalent resistance (R_{eq}) for resistors in series is given by:

R{eq} = R1 + R2 + R3 + …

Voltage Divider

A voltage divider is a series circuit used to produce a source of potential difference that is less than the potential difference across the battery.
Voltage dividers are often used with sensors, such as photoresistors.
The resistance of a photoresistor depends on the amount of light that strikes it; it can be used in a light meter.
In this device, the potential difference is converted to a measurement of illuminance.

Series Circuit Example Problem 1

Problem: Suppose 15 V are applied across three resistors in series (15.0 \Omega , 22.0 \Omega , and 47.0 \Omega ). Determine the current in the circuit and the potential difference across the 47.0- \Omega resistor.
- Known:
  - V_{source} = 15 V
  - R_1 = 15.0 \Omega
  - R_2 = 22.0 \Omega
  - R_3 = 47.0 \Omega
- Unknown:
  - I = ?
  - V_3 = ?
- Solution:
  1. Determine the equivalent resistance:
    R{eq} = R1 + R2 + R3 = 15.0 \Omega + 22.0 \Omega + 47.0 \Omega = 84.0 \Omega
  2. Use the relationship among potential difference, current, and total resistance to find the current through the circuit and the potential difference across R3.
    I = \frac{V{source}}{R{eq}} = \frac{15 V}{84.0 \Omega} = 0.18 A
    V3 = I \cdot R3 = (0.18 A)(47.0 \Omega) = 8.5 V
The units are correct (amperes for current and volts for potential difference).

Series Circuit Example Problem 2

Problem: Demonstrate that the total power is equal to the sum of the individual power amounts used in the three resistors.
- Known:
  - V_{source} = 15 V
  - I = 0.18 A
  - R_1 = 15.0 \Omega
  - R_2 = 22.0 \Omega
  - R_3 = 47.0 \Omega
- Unknown:
  - P = ?
  - P_1 = ?
  - P_2 = ?
  - P_3 = ?
- Solution:
  1. Use the relationship among power, potential difference, and current to find the power for the whole circuit.
    P = V_{source} \cdot I = (15 V)(0.18 A) = 2.7 W
  2. Use the relationship among power, current, and resistance for each resistor.
    P1 = I^2 \cdot R1 = (0.18 A)^2 (15.0 \Omega) = 0.49 W
    P2 = I^2 \cdot R2 = (0.18 A)^2 (22.0 \Omega) = 0.71 W
    P3 = I^2 \cdot R3 = (0.18 A)^2 (47.0 \Omega) = 1.52 W
  3. Add the individual powers:
    P{total} = P1 + P2 + P3 = 0.49 W + 0.71 W + 1.52 W = 2.72 W
The total power equals the sum of the power for the individual resistors, as expected.

Series Circuits Example Problem 3

Problem: A voltage divider consisting of two 1.5-M \Omega resistors is connected to a 12.0-V source. Determine the potential difference across each resistor before and after a voltmeter is connected, assuming the voltmeter’s resistance is 1.0×10^7 \Omega .
- Known:
  - V_{source} = 12.0 V
  - R1 = R2 = 1.5 \times 10^6 \Omega
  - R_V = 1.0 \times 10^7 \Omega
- Unknown:
  - V_1, before = ?
  - V_2, before = ?
  - V_1, after = ?
  - V_2, after = ?
- Solution:
  1. Before the voltmeter is connected, the potential difference across each resistor will be half the supply potential difference, or 6.0 V.
  2. When the voltmeter is connected, it acts as a parallel resistance.
  3. Find the voltage drop across the parallel combination.
  4. Use the loop rule to find the potential difference across the first resistor.

Parallel Circuits

A circuit in which there are several current paths is called a parallel circuit.
In the mountain river model, such a circuit is illustrated by multiple paths for the water over a waterfall.
Some paths might have a large flow of water, while others might have a small flow.
The sum of the flows, however, is equal to the total flow of water over the falls.
In addition, regardless of which channel the water flows through, the drop in height is the same.
Similarly, in a parallel electric circuit, the total current is the sum of the currents through each path, and the potential difference across each path is the same.
The current through each resistor in a parallel electric circuit depends upon the individual resistances.
The branches of a parallel circuit are independent of each other. The current through each resistor depends only upon the potential difference across it and its resistance.
The total current is the sum of the current in the branches.
I{total} = I1 + I2 + I3 + …
The total current is also equal to the potential difference across the source divided by the equivalent resistance (R).
I = \frac{V}{R}
Placing more resistors in parallel always decreases the equivalent resistance of a circuit. The resistance decreases because each new resistor provides an additional path for current, thereby increasing the total current while the potential difference remains unchanged.

Equivalent Resistance for Resistors in Parallel

The equivalent resistance for resistors in parallel is given by:
\frac{1}{R{eq}} = \frac{1}{R1} + \frac{1}{R2} + \frac{1}{R3} + …

Parallel Circuits Example Problem

Problem: Four resistors, 50.0 \Omega , 40.0 \Omega , 30.0 \Omega , and 20.0 \Omega , are connected with a parallel circuit across a 120.0-V battery. Find the current through each branch of the circuit, the equivalent resistance of the circuit, and the current through the battery.
- Known:
  - V_{battery} = 120.0 V
  - R_A = 50.0 \Omega
  - R_B = 40.0 \Omega
  - R_C = 30.0 \Omega
  - R_D = 20.0 \Omega
- Unknown:
  - I_{battery} = ?
  - R_{eq} = ?
  - I_A = ?
  - I_B = ?
  - I_C = ?
  - I_D = ?
- Solution:
  1. The potential difference is the same across all four resistors.
  2. Use the relationship among potential difference, current, and resistance:
    IA = \frac{V{battery}}{RA} = \frac{120.0 V}{50.0 \Omega} = 2.4 A IB = \frac{V{battery}}{RB} = \frac{120.0 V}{40.0 \Omega} = 3.0 A
    IC = \frac{V{battery}}{RC} = \frac{120.0 V}{30.0 \Omega} = 4.0 A ID = \frac{V{battery}}{RD} = \frac{120.0 V}{20.0 \Omega} = 6.0 A
  3. Determine the equivalent resistance:
    \frac{1}{R{eq}} = \frac{1}{50.0 \Omega} + \frac{1}{40.0 \Omega} + \frac{1}{30.0 \Omega} + \frac{1}{20.0 \Omega} R{eq} = 7.79 \Omega
  4. The current through the battery is equal to the potential difference across the battery divided by the equivalent resistance.
    I{battery} = \frac{V{battery}}{R_{eq}} = \frac{120.0 V}{7.79 \Omega} = 15.4 A
The sum of the currents in the branches equals the current through the battery.

Kirchhoff's Rules

Gustav Robert Kirchhoff was a German physicist who formulated two rules that govern electric circuits:
- The loop rule
- The junction rule

Loop Rule

Describes electric potential differences and is based on the law of conservation of energy.
It states that the sum of increases in electric potential around a loop in an electric circuit equals the sum of decreases in electric potential around that loop.
For an application, picture an electric current traveling clockwise around the red loop in the figure.
Electric current traveling clockwise around the red loop in the figure, electric potential increases by 9V as this charge travels through the battery, and electric potential drops by 5V as this charge travels through resistor 1.
By the loop rule, the increases in electric potential around a loop must equal the decreases in electric potential around that loop. Therefore, the drop in electric potential across resistor 2 must be 9V – 5V = 4V.
Note that resistor 3 does not affect our answer because resistor 3 is not a part of the loop that includes the battery, resistor 1, and resistor 2.

Junction Rule

Describes currents and is based on the law of conservation of charge.
Recall the law of conservation of charge states that charge can neither be created or destroyed.
This means that, in an electric circuit, the total current into a section of that circuit must equal the total current out of that same section.
A junction is a location where three or more wires are connected together.
According to Kirchhoff’s rule, the sum of currents entering a junction is equal to the sum of currents leaving that junction. Otherwise, charge would build up at the junction.
In the figure, I1 = I2 + I3 at junction A, and I2 + I3 = I1 at junction B.