Question

Answer the following giving reason :
A low voltage supply from which one needs high currents must have very low internal resistance. Why ?

Hint:

Car batteries are the perfect real-world example of this exact principle. They are only 12 Volts (relatively low), but they need to supply hundreds of Amperes to start an engine. Therefore, car batteries are heavily engineered to have an internal resistance of just a few milliohms.

Answer 1

Concept:
• Any real-world voltage supply (like a battery) consists of an ideal electromotive force (emf, $E$) strictly in series with an unavoidable internal resistance ($r$).
• When current flows out of the battery, a specific portion of the generated voltage is inevitably lost inside the battery itself due to this internal resistance.
• The macroscopic terminal voltage equation is given by $V = E - Ir$.

Step 1: Analyze the maximum current formula
According to Ohm's law applied to a complete circuit, the total current $I$ drawn from a battery connected to an external load resistance $R$ is strictly formulated as:
\[ I = \frac{E}{R + r} \]
To physically draw the absolute maximum theoretical current from this specific supply, we must short-circuit the terminals, making the external resistance absolutely zero ($R = 0$).
The maximum possible current equation then rigorously simplifies to:
\[ I_{max} = \frac{E}{r} \]

Step 2: Evaluate the specific given condition
The problem specifically states that we are utilizing a "low voltage supply."
This means the numerator in our equation, the electromotive force $E$, is a remarkably small numerical value.
Simultaneously, the problem strictly requires this weak supply to deliver "high currents."
This mathematically demands that the overall resulting fraction ($E/r$) must be an exceptionally large number.

Step 3: Determine the necessary condition for internal resistance
In any fraction where the numerator is fixed to a small value, the only mathematically valid way to make the total quotient extremely large is to make the denominator incredibly tiny.
Therefore, the internal resistance $r$ strictly situated in the denominator must be extremely low.

Step 4: Conclusion
If the internal resistance $r$ were high, the substantial voltage drop strictly occurring inside the battery itself ($V_{drop} = Ir$) would completely consume the already low available emf $E$.
Thus, to prevent massive internal voltage loss and successfully deliver a high current, a low voltage supply must inherently be constructed with a very low internal resistance.