Grade 10

Advanced

8 min

Lesson 6 of 12

Parallel Circuit Analysis

Not Started

Examines circuits with multiple paths and how current divides among branches.

Why is it that when one light bulb in your house burns out, the rest of the room doesn't go dark? The secret lies in the 'choose your own adventure' logic of parallel circuits.

Learning Objectives

1.Calculate the equivalent resistance of resistors in parallel using the reciprocal formula
2.Explain why the potential difference remains identical across all branches
3.Analyze how adding parallel branches increases total current while decreasing total resistance

The Logic of Multiple Paths

In a parallel circuit, components are connected across the same two nodes, creating multiple independent paths for the electric current. Think of it like a highway that suddenly expands from one lane to three; the cars (electrons) can choose which lane to take. The most fundamental rule of parallel circuits is that the potential difference ( $V$ ) is the same across every branch. Because every branch is connected directly to the same two terminals of the power source, the 'electrical pressure' pushing the charges is identical, regardless of the resistance in that specific path. This is why all appliances in your home are wired in parallel—they all need the same standard voltage to operate correctly.

Quick Check

If a 12V battery is connected to three resistors in parallel, what is the voltage across the third resistor?

Recall the 'highway' analogy where every lane starts and ends at the same point.

Answer

The voltage is 12V because potential difference is the same across all branches in a parallel circuit.

The Reciprocal Resistance Paradox

Calculating the equivalent resistance (

R_p

) in parallel is often counter-intuitive. In a series circuit, adding resistors increases total resistance. However, in parallel, adding more resistors actually decreases the total resistance. This is because you are providing more 'lanes' for the current to flow through, making it easier for the total charge to move from one side of the battery to the other. We use the reciprocal formula to find the total resistance:

\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \dots + \frac{1}{R_n}

After adding the fractions, you must take the reciprocal of the result to find

R_p

Calculate the equivalent resistance of a $10\Omega$ and a $15\Omega$ resistor connected in parallel.

1. Write the formula: $\frac{1}{R_p} = \frac{1}{10} + \frac{1}{15}$ 2. Find a common denominator (30): $\frac{1}{R_p} = \frac{3}{30} + \frac{2}{30} = \frac{5}{30}$ 3. Simplify the fraction: $\frac{1}{R_p} = \frac{1}{6}$ 4. Take the reciprocal: $R_p = 6\Omega$

*Observation: Notice that $6\Omega$ is smaller than both individual resistors.*

Quick Check

True or False: The equivalent resistance of a parallel circuit is always less than the resistance of the smallest individual resistor in the circuit.

Think about adding an extra door to a crowded room; does it make it easier or harder to leave?

Answer

True. Adding a path always reduces the overall difficulty for current to flow.

Current Division and Kirchhoff's Law

While voltage is constant, the total current (

I_{total}

) is the sum of the currents in each individual branch. This is known as Kirchhoff's Junction Rule, which states that the charge entering a junction must equal the charge leaving it:

I_{total} = I_1 + I_2 + \dots + I_n

Using Ohm's Law (

I = \frac{V}{R}

), we can see that branches with lower resistance will draw more current. This is why plugging too many devices into one parallel power strip can be dangerous; each device adds a new path, increasing the total current drawn from the wall until it potentially exceeds the safety limit of your wires.

A circuit has a $24V$ source and two parallel branches. Branch 1 has a $12\Omega$ resistor. The total current leaving the battery is $5A$ . Find the resistance of the second branch ( $R_2$ ).

1. Find the current in Branch 1: $I_1 = \frac{V}{R_1} = \frac{24}{12} = 2A$ 2. Use the current sum rule to find $I_2$ : $I_2 = I_{total} - I_1 = 5A - 2A = 3A$ 3. Use Ohm's Law to find $R_2$ : $R_2 = \frac{V}{I_2} = \frac{24}{3} = 8\Omega$ 4. Verify: $\frac{1}{R_p} = \frac{1}{12} + \frac{1}{8} = \frac{2+3}{24} = \frac{5}{24}$ . $R_p = 4.8\Omega$ . $I_{total} = \frac{24}{4.8} = 5A$ . Correct!

Key Takeaways

1Voltage ( $V$ ) is identical across all branches of a parallel circuit.
2The total current ( $I_{total}$ ) is the sum of all individual branch currents.
3Adding more parallel branches decreases the total equivalent resistance ( $R_p$ ).
4The reciprocal formula is: $\frac{1}{R_p} = \sum \frac{1}{R_i}$ .

Quick Check

Multiple Choice

What happens to the total resistance of a circuit if you add a third resistor in parallel to two existing ones?

Multiple Choice

In a parallel circuit with a $20\Omega$ and a $60\Omega$ resistor, which resistor will have the higher current?

True/False

If one branch in a parallel circuit is broken (open), the current in the other branches will stop flowing.

What's Next?

Review Tomorrow

In 24 hours, try to sketch a 3-branch parallel circuit and write down the formula for equivalent resistance from memory.

Practice Activity

Look at the 'Total Amps' rating on a power strip in your house. Research why 'daisy-chaining' multiple power strips is a fire hazard based on the parallel current rule.

Browse

Browse

Parallel Circuit Analysis

Learning Objectives

The Logic of Multiple Paths

The Reciprocal Resistance Paradox

Current Division and Kirchhoff's Law

Key Takeaways

Quick Check

What's Next?

Parallel Circuit Analysis

Learning Objectives

The Logic of Multiple Paths

The Reciprocal Resistance Paradox

Current Division and Kirchhoff's Law

Key Takeaways

Quick Check

What's Next?