Question

A low voltage supply from which one needs high currents must have very low internal resistance. Why?

Answer 1

The Simple Analogy: A Water Pipe

Imagine your power supply is a large water tank (representing the voltage) and you need to get a high flow of water (high current) out of a pipe at the bottom.

• Internal Resistance is like a narrow, clogged section or a pinch inside that pipe.
• Low Internal Resistance is a wide, clear pipe.

If you need a high flow of water (high current), the wide, clear pipe (low internal resistance) will let the water rush out with very little loss of pressure. The pressure at the end of the pipe will be almost the same as the pressure in the tank.

However, if you try to force that same high flow of water through the narrow, pinched pipe (high internal resistance), you'll have a big problem. The pinch itself creates a massive pressure drop. The pressure (voltage) available at the end of the pipe will be significantly lower than the pressure in the tank.

For a low-pressure tank (low voltage supply) to begin with, this pressure drop can be so severe that you get barely a trickle out of the end, even though the tank is full.

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The Technical Explanation (Using Ohm's Law)

Every real-world power supply can be modeled as an ideal voltage source in series with an internal resistor (r_int).

Here's what the components mean:
V_supply (or EMF): The ideal, "no-load" voltage of the supply. This is the voltage you'd measure if nothing was connected to it.
r_int: The internal resistance of the supply. This represents the resistance of the battery's chemicals, the windings of a transformer, etc.
R_load: The resistance of the device you are powering (the "load").
I: The current flowing through the entire circuit.
* V_terminal: The actual voltage measured at the terminals of the supply when the device is connected and running. This is the voltage your device actually "sees".

The key equation comes from analyzing this circuit:

*V_terminal = V_supply - (I r_int)**

This equation tells us that the terminal voltage (what the device gets) is the ideal supply voltage minus a voltage drop that occurs inside the power supply itself. This internal voltage drop (I * r_int) is directly proportional to both the current being drawn and the internal resistance.

Why This Matters for Low Voltage, High Current Supplies

Let's look at two scenarios to see why a low r_int is crucial.

Scenario 1: BAD Design (High Internal Resistance)
A low-voltage supply: V_supply = 5V
A high internal resistance: r_int = 1 Ω
* We need to power a device that needs a high current: I = 4A

1. Calculate the internal voltage drop:
   V_drop = I * r_int = 4A * 1Ω = 4V

2. Calculate the actual terminal voltage the device sees:
   V_terminal = V_supply - V_drop = 5V - 4V = 1V

Result: Even though you have a 5V supply, it can only deliver 1V to your device when it tries to draw 4A. Your device will almost certainly fail to operate. A huge amount of the supply's energy is being wasted as heat inside the supply itself.

Scenario 2: GOOD Design (Low Internal Resistance)
A low-voltage supply: V_supply = 5V
A very low internal resistance: r_int = 0.05 Ω
* We need to power the same device: I = 4A

1. Calculate the internal voltage drop:
   V_drop = I * r_int = 4A * 0.05Ω = 0.2V

2. Calculate the actual terminal voltage the device sees:
   V_terminal = V_supply - V_drop = 5V - 0.2V = 4.8V

Result: The supply successfully delivers 4.8V to the device. This is very close to its ideal voltage, and the device will operate correctly.

Summary of Key Reasons

1. To Minimize Voltage Drop (Voltage Sag): High current multiplied by even a small internal resistance causes a voltage drop. If the starting voltage is already low, this drop represents a significant percentage of the total voltage, starving the load.

2. To Maximize Power Transfer: A power supply delivers maximum power to a load when the load's resistance matches the supply's internal resistance. However, in this state, 50% of the power is wasted as heat inside the supply! For efficient power delivery, you want the internal resistance to be a tiny fraction of the load's resistance.

3. To Prevent Overheating: The power wasted as heat inside the supply is calculated by *P_waste = I² r_int**. Since the current (I) is high, squaring it makes it a very large number. If r_int is not extremely small, this will generate a massive amount of heat, which can damage or destroy the power supply.