Electrical Engineering Calculator

Voltage Divider Calculator

Calculate output voltage, divider current, resistor power dissipation, and total power for an ideal unloaded two-resistor voltage divider.

Circuit Diagram

Standard unloaded voltage divider network

Vin = 12 VR1 = 1000 ΩR2 = 2000 ΩVout

Circuit Parameters

V

Total source voltage supplied to the divider.

Ω

Upper resistor connected to the input source.

Ω

Lower resistor connected to ground. Vout is measured across R2.

Material Note

Material properties can vary significantly based on heat treatment and manufacturer. Always consult specific supplier datasheets for critical components.

Output Voltage (Vout)

8.00V

Electrical Results

Divider Current

4.00 mA

Current through R1 and R2

R1 Power

16.00 mW

Power dissipated by R1

R2 Power

32.00 mW

Power dissipated by R2

Total Power

48.00 mW

Total power drawn from the source

Governing Formula

V_out = V_in · [ R2 / (R1 + R2) ]
V_outV

Output voltage across R2

V_inV

Input source voltage

R1Ω

Upper resistance

R2Ω

Lower resistance

Governing Formula

I = V_in / (R1 + R2)
IA

Divider current

V_inV

Input source voltage

R1Ω

Upper resistance

R2Ω

Lower resistance

Calculation Assumptions

  • Ideal voltage source
  • No source internal resistance
  • Unloaded Vout
  • Linear resistors
  • DC steady-state operation
  • No parallel load at Vout

Engineering Code

Equivalent implementation of the ideal unloaded voltage divider calculation.

Python
def voltage_divider(v_in, r1, r2):
    """
    Calculate an ideal unloaded voltage divider.

    v_in: Input voltage (V)
    r1: Upper resistor (Ohms)
    r2: Lower resistor (Ohms)
    """

    if not all(map(lambda x: x == x and abs(x) != float("inf"), [v_in, r1, r2])):
        raise ValueError("Inputs must be finite numbers.")

    if r1 <= 0 or r2 <= 0:
        raise ValueError("Resistors must be greater than zero.")

    current = v_in / (r1 + r2)
    v_out = v_in * (r2 / (r1 + r2))

    p_r1 = current**2 * r1
    p_r2 = current**2 * r2
    p_total = v_in * current

    return v_out, current, p_r1, p_r2, p_total


# Example Usage
Vin = 12
R1 = 1000
R2 = 2000

Vout, I, P_R1, P_R2, P_total = voltage_divider(Vin, R1, R2)

print(f"Output Voltage: {Vout:.2f} V")
print(f"Divider Current: {I * 1000:.2f} mA")
print(f"R1 Power: {P_R1 * 1000:.2f} mW")
print(f"R2 Power: {P_R2 * 1000:.2f} mW")
print(f"Total Power: {P_total * 1000:.2f} mW")
MATLAB
function [v_out, current, p_r1, p_r2, p_total] = voltage_divider(v_in, r1, r2)
    % Calculate an ideal unloaded voltage divider.
    %
    % v_in  - Input voltage (V)
    % r1    - Upper resistor (Ohms)
    % r2    - Lower resistor (Ohms)

    if ~isfinite(v_in) || ~isfinite(r1) || ~isfinite(r2)
        error('All inputs must be finite numbers.');
    end

    if r1 <= 0 || r2 <= 0
        error('Resistor values must be strictly positive.');
    end

    current = v_in / (r1 + r2);
    v_out = v_in * (r2 / (r1 + r2));

    p_r1 = current^2 * r1;
    p_r2 = current^2 * r2;
    p_total = v_in * current;
end

% Example Usage
Vin = 12;
R1 = 1000;
R2 = 2000;

[Vout, I, P_R1, P_R2, P_total] = voltage_divider(Vin, R1, R2);

fprintf('Output Voltage: %.2f V\n', Vout);
fprintf('Divider Current: %.2f mA\n', I * 1000);
fprintf('R1 Power: %.2f mW\n', P_R1 * 1000);
fprintf('R2 Power: %.2f mW\n', P_R2 * 1000);
fprintf('Total Power: %.2f mW\n', P_total * 1000);

Technical Explanation: Voltage Dividers

A voltage divider is a two-resistor network used to produce an output voltage that is a fraction of the input voltage. For an ideal unloaded divider, the output voltage is determined by the ratio of R1 and R2.

Voltage Divider Equation

With R1 connected between the source and Vout, and R2 connected between Vout and ground, the output voltage is:

Vout = Vin × R2 / (R1 + R2)

Divider Current

In the unloaded condition, the same current flows through R1 and R2. The divider current is:

I = Vin / (R1 + R2)

Resistor Power Dissipation

Each resistor dissipates power according to P = I²R. Therefore:

PR1 = I²R1
PR2 = I²R2

The total power drawn from the source is the sum of the resistor losses:

Ptotal = Vin × I = Vin² / (R1 + R2)

Loading Effect

The standard voltage divider equation assumes that no significant current is drawn from the Vout terminal. If a load resistance RL is connected across R2, the effective lower resistance becomes:

R2,eff = R2 || RL = (R2 × RL) / (R2 + RL)

The loaded output voltage must then be calculated using R2,eff instead of R2. Because the parallel combination is always lower than R2 for a finite positive RL, the loaded output voltage is lower than the ideal unloaded value.

Practical Design Considerations

A voltage divider is generally suitable for reference voltages, bias networks, sensing circuits, and high-impedance inputs such as ADCs. It is generally unsuitable as a power supply for loads that draw significant or variable current, because the output voltage changes with loading and power is continuously dissipated in the resistors.

In a practical design, the calculated power dissipation of each resistor should be compared with its rated power, with an appropriate design margin for temperature and operating conditions.

How to Use the Voltage Divider Calculator

  1. Source Voltage (Vin): Enter the initial input voltage supplied to the circuit.
  2. Resistor 1 (R1): Input the value of the first (top) resistor.
  3. Resistor 2 (R2): Input the value of the second (bottom) resistor connected to the ground.
  4. Output Voltage (Vout): The calculator determines the exact stepped-down voltage across R2.

Real-World Engineering Cases

The Microcontroller Brownout Crash

A hobbyist used a simple resistor voltage divider to step down a 12V battery supply to 5V in order to power an expensive microcontroller. As soon as the microcontroller initiated a high-power WiFi transmission, its internal resistance dropped, effectively placing a new load in parallel with R2. This altered the divider ratio, causing the voltage to plummet to 2V. The MCU browned out and corrupted its own flash memory.

Engineering Lesson

Never use an unbuffered voltage divider to provide power to a fluctuating load. Voltage dividers are strictly for stepping down high-impedance reference signals (like ADC inputs). For power delivery, always use a linear regulator or a buck converter.

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Frequently Asked Questions

Can I use a voltage divider as a power supply?

A voltage divider is generally unsuitable as a power supply for loads that draw significant or variable current. The load changes the effective resistance of the divider and therefore changes Vout. Voltage dividers are better suited to reference, bias, and sensing applications with sufficiently high input impedance.

What happens if I connect a load resistor across R2?

The load resistance RL becomes parallel with R2. The effective resistance is R2 || RL, which is lower than R2 for any finite positive RL. Consequently, the actual output voltage becomes lower than the unloaded voltage-divider value.

What happens if I use very high resistor values?

Higher resistance values reduce divider current and power consumption. However, they also increase the output impedance, making Vout more sensitive to loading, leakage currents, measurement equipment, and electrical noise.

How do I calculate resistor power?

For an unloaded divider, the same current flows through R1 and R2. Resistor power can therefore be calculated using P = I²R. The total divider power is Ptotal = Vin² / (R1 + R2).

Does the calculator account for resistor tolerance?

No. This calculator uses the nominal resistance values and assumes ideal resistors. Actual Vout can vary because of resistor tolerance, temperature coefficient, source resistance, and loading effects.