How does a load decide the amount of current drawn in a circuit?

Question

I realise this question might seem similar to other questions asked on the site. But what I'm asking is actually completely different.

Suppose I have a 5ohm resistive load hooked up to a 10v battery. The current drawn by the load would be equal to the current through the circuit, right?(assuming the load is connected in series). Using ohm's law, the current through the circuit would hence be 2A.

Now,if I connect the same 5ohm resistive load to a 15V battery, the current through the circuit would be 3A, using ohm's law.

So doesn't the same load draw different currents when the supply voltage changes? If that's the case, why do we say "current drawn depends on the type of load connected"?

Moreover does the above reasoning of using ohm's law hold valid for any type of load? (i.e. L load, C load, RL load or a combination of all 3)

Edit: this highly upvotedsays that for a constant voltage supply, current drawn depends on load. So how does the load draw two different currents for two different voltages?

Answer 1

why do we say "current drawn depends on the type of load connected

We actually say that the current drawn depends on the load AND the power supply voltage. In other words, what we should say matches ohms law.

Moreover does the above reasoning of using ohm's law hold valid for any type of load? i.e. L load, C load, RL load or a combination of all 3)

No, it becomes more complex with capacitors and inductors because the current drawn by a capacitor depends on the rate of change of applied voltage and not the absolute value of voltage.

For an inductor, the current depends on the integral of voltage with respect to time hence, inductor current is proportional to applied voltage x time.

That’s for DC circuits. It’s another departure from ohms law for AC circuits and handling complex transients but the same rate of change (or integral) principles apply.

Answer 2

If that's the case, why do we say "current drawn depends on the type of load connected"?

Usually we're dealing with a fixed / constant voltage supply. Most common examples would be domestic or industrial mains voltages or 12 V automotive systems. In both cases the voltage remains within certain tolerances so that means the other's have an inverse relationship, \$ I = \frac {V}{R} \$ or, since V is constant, \$ I \propto \frac {1}{R} \$.

Figure 1. Each of these various bulbs, connected to the same supply, will present a different effective resistance and will draw a current inversely proportional to their resistance. The higher the power consumption (watts) the lower the effective 'resistance'. (I use the word resistance cautiously here as LEDs are not resistances but it's getting harder to find images of incandescent lamps!) Image source: Banggood.

Moreover does the above reasoning of using ohm's law hold valid for any type of load? (i.e. L load, C load, RL load or a combination of all three).

No. Ohm's law is specific to resistance. We can extend the law to inductors and capacitors but we have to calculate impedances and use complex numbers.

Edit: this highly upvoted says that for a constant voltage supply, current drawn depends on load. So how does the load draw two different currents for two different voltages?

Because you've changed the voltage. There's no inconsistency there.

Answer 3

It depends both on supplied voltage but also type of load. Resistive load draws current based on Ohm's law so it depends only on supplied voltage and the load resistance. Then, there are devices that have linear regulators, so they internally convert everything to 5V for example so with a fixed load after regulator, they basically draw constant current regardless of input voltage. Then there are devices with switch mode regulators so with a fixed load after regulator, they always draw constant power, so when voltage is increased, they draw less current.

Answer 4

All conductors , semiconductors, capacitors and inductors have resistance even regulated power supplies.

• Regulated supplies amplify the error between desired output and actual but response time is bandwidth or slew rate limited.
• Thus load regulation error appears in all regulator specs
• It can be defined a number of ways.
• 1st order estimate Zout= ΔV/ΔI[Ω] above some minimal load to avoid other issues.
  • In Batteries and caps, it's called ESR.
• 2nd order impedance is the storage capacitance which limits slew rate on errors. On Batteries, they recover quickly from a brief short. This memory effect is due to another effect from Double-charge layer effects which are modelled as another RC with bigger C and R that remembers what the brief short circuit that the discharge depleted.

Let's look at a simple model.

^{simulate this circuit – Schematic created using CircuitLab}

I have discussed the models from memory effects of batteries elsewhere, if you are interested.

We use many different symbols for the internal resistance, Rs, Ri, ESR, \$r_{be}, R_{ce}\$ and for inductors, DCR and for Zeners, Zzt.

---

Let's look at one of the cheapest LDO's ($0.05653 on 3k reel)

recall Zout= ΔV/ΔI[Ω]

here Zout= ΔV_{out}/ΔI = 30 mV / 150 mA = 200 mΩ

Notice errors include Vout tolerance over temp , Line REG and Load REG.

Also note the fast rise time in the spec. This tells you slower loads may be regulated better due to high gain , but this has an internal rise time limit to respond, which is why external Low ESR caps are needed to reduce any source of high freq. ripple.