Voltage drop in a parallel circuit refers to the decrease in voltage as a charge travels through the components in the circuit. In a parallel circuit, the voltage is equal across all components, meaning that the voltage drop across each component is the same. This is because the components in a parallel circuit are connected across the same pair of nodes, and the voltage measured between sets of common points must always be the same at any given time.

## Understanding Voltage Drop in Parallel Circuits

The voltage drop across each resistor in a parallel circuit can be calculated using Ohm’s law, which states that the voltage drop (V) across a resistor is equal to the current (I) flowing through the resistor multiplied by the resistance (R) of the resistor (V = IR). Since the voltage is the same across all components in a parallel circuit, the current through each resistor will be proportional to its resistance (I = V/R).

For example, consider a parallel circuit with three resistors (R1, R2, and R3) connected across a battery with a voltage of V. The current through each resistor (I1, I2, and I3) can be calculated using Ohm’s law (I1 = V/R1, I2 = V/R2, and I3 = V/R3), and the total current (Itot) in the circuit can be found by adding the currents through all the branches (Itot = I1 + I2 + I3).

### Theorem: Voltage Drop in a Parallel Circuit

The theorem states that the voltage drop in a parallel circuit is the same across all components. This is because the components in a parallel circuit are connected across the same pair of nodes, and the voltage measured between sets of common points must always be the same at any given time.

### Electronics Formula: Voltage Drop

The voltage drop (V) across a resistor in a parallel circuit can be calculated using the formula:

V = I × R

Where:

– V is the voltage drop (in volts)

– I is the current flowing through the resistor (in amperes)

– R is the resistance of the resistor (in ohms)

### Electronics Example: Parallel Circuit with Three Resistors

Consider a parallel circuit with three resistors (R1 = 10 kΩ, R2 = 2 kΩ, and R3 = 1 kΩ) connected across a battery with a voltage of 9 V. The current through each resistor can be calculated as follows:

- I1 = 9 V / 10 kΩ = 0.9 mA
- I2 = 9 V / 2 kΩ = 4.5 mA
- I3 = 9 V / 1 kΩ = 9 mA

The total current in the circuit is Itot = I1 + I2 + I3 = 0.9 mA + 4.5 mA + 9 mA = 14.4 mA.

## Electronics Numerical Problem: Parallel Circuit with Three Resistors

A parallel circuit has three resistors with resistance values of 10 Ω, 20 Ω, and 30 Ω. The voltage across the circuit is 100 V. Calculate the voltage drop across each resistor and the total current in the circuit.

**Data Points/Values:**

– Voltage across the circuit = 100 V

– R1 = 10 Ω

– R2 = 20 Ω

– R3 = 30 Ω

**Measurements:**

– Voltage drop across R1 = 100 V / (10 Ω + 20 Ω + 30 Ω) × 10 Ω = 14.29 V

– Voltage drop across R2 = 14.29 V

– Voltage drop across R3 = 14.29 V

– Total current in the circuit = 100 V / (10 Ω + 20 Ω + 30 Ω) = 2.38 A

## Conclusion

In summary, voltage drop in a parallel circuit is the decrease in voltage as a charge travels through the components in the circuit, and the voltage drop across each component is the same. This can be calculated using Ohm’s law, and the total current in the circuit can be found by adding the currents through all the branches.

## References

- Parallel Circuits and the Application of Ohm’s Law, All About Circuits, https://www.allaboutcircuits.com/textbook/direct-current/chpt-5/simple-parallel-circuits/
- Physics Tutorial: Parallel Circuits, Physics Classroom, https://www.physicsclassroom.com/class/circuits/Lesson-4/Parallel-Circuits
- Can someone explain voltage drop (in parallel) to me?, Reddit, https://www.reddit.com/r/Mcat/comments/ru8m8g/can_someone_explain_voltage_drop_in_parallel_to_me/

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