What if a battery had more voltage?

A battery makes a bulb light up. It pushes electricity around a loop of wire.

Some batteries push harder than others. What if the battery pushed much harder?

A battery has a number printed on it, like 1.5 V or 9 V. The V stands for volts, and the number tells you the battery's voltage: how hard it pushes electric current around a circuit.

In this lab, Olly has two small bulbs joined in one loop with a 3-volt battery. What would happen if he swapped in a battery with three times the voltage? Would the bulbs get the same current, more current, or less?

Voltage, or potential difference, is measured in volts and drives a current through a resistance. It also sets how much electrical energy the moving charge carries: OpenStax writes that energy as PE = qV, the charge moved times the voltage.

This lab holds the circuit fixed, two identical 3 Ω bulbs in series with an ideal battery, and changes only the source voltage, from 3 V to 9 V. The question is how the current, each bulb's power and the energy used scale with that one change.

Olly starts with a small battery. It pushes a little bit of electricity through two bulbs.

Both bulbs glow, but only softly.

With the 3-volt battery, the two bulbs share the push: each one gets 1.5 volts. The current, the flow of electricity, is 0.50 amperes (amps for short), and it is the same all the way around the loop.

Each bulb turns 0.75 watts of electrical power into light and heat, so the battery supplies 1.5 watts in all. In the model's 12-second run, that adds up to 18 joules of energy.

Two 3 Ω bulbs in series make 6 Ω, so Ohm's law gives I = 3 V ÷ 6 Ω = 0.50 A through the whole loop. Each bulb drops 1.50 V and dissipates P = I²R = 0.75 W; the source delivers 1.5 W, or 18 J over the model's 12-second run.

The model's brightness meter reads 25% for each bulb, because it compares each bulb's power with full glow: a fixed 3 W, what one bulb alone gets from 3 V. That yardstick stays the same when the voltage changes.

Now Olly uses a battery that pushes three times as hard.

Three times as much electricity flows. Each bulb gets much more energy every second!

Turn the battery up to 9 volts and keep the same two bulbs. The current triples, from 0.50 to 1.50 amps: watch the moving dots speed up and the ammeter's reading climb. Each bulb now gets 4.5 volts.

The power grows even faster. Each bulb now uses 6.75 watts, nine times as much as before, and the battery supplies 13.5 watts. In the 12-second run, the energy used climbs from 18 to 162 joules. Three times the push gives three times the current, and nine times the power.

At 9 V, I = 9 V ÷ 6 Ω = 1.50 A, three times the 3 V current, and each bulb drops 4.50 V. Power scales with the square: P = I²R = (1.50 A)² × 3 Ω = 6.75 W per bulb, nine times 0.75 W, with 13.5 W from the source and 162 J over 12 seconds.

The brightness meter climbs from 25% to 225%, nine times as much, because it measures every bulb against the same fixed 3 W full glow, and the drawn bulbs turn white-hot. At more than twice full glow, the model notes that a real bulb would soon burn out; its own bulbs never do. Try 12 V: 2.00 A, 12 W and 400% per bulb.

Voltage is the battery's push. A bigger push moves more electricity through the bulbs.

More electricity, pushed harder, gives each bulb lots more energy. That is why the power grows so fast.

The bulbs resist the flow of current. A bigger push gets more current through the same resistance: that rule is called Ohm's law. Three times the voltage gives three times the current.

Power depends on both the push and the flow. Each bulb now gets three times the voltage and three times the current, and three times three is nine. Our model keeps things simple: its bulbs always resist the same amount, its battery never runs down, and its bulbs never burn out, however much power they get.

Ohm's law, I = V/R, makes current proportional to voltage when the resistance is fixed. Electrical power is P = IV = V²/R = I²R, so with R constant, power goes as the square of the voltage: tripling V multiplies P by nine. OpenStax puts it this way: because voltage is squared, raising it has a bigger effect than you might expect.

The model's limits matter here. Its bulbs are fixed 3 Ω resistors, but a real tungsten filament's resistance rises as it heats (OpenStax works an example from 0.350 Ω cold to 4.8 Ω at 2,850 °C). Its battery holds a perfectly steady voltage, its bulbs never burn out, and brightness is compared by power, not by light.

Real batteries have their voltage printed on them. A small AA battery says 1.5 volts.

A bulb is made for one push. Too big a push can make a real bulb burn out!

Battery makers print the voltage on each battery. A dry cell gives about 1.5 volts whatever its size, so AA, AAA, C and D cells all push the same; bigger ones just last longer. A 9-volt battery is labeled 9 volts. Flashlights and toys often put cells in a row, which adds their voltages together: two 1.5-volt cells make 3 volts.

A real bulb is built for one voltage. The OpenStax physics textbook describes a 25-watt bulb made for 120 volts that is plugged into 240 volts: it glows very brightly for a moment, then burns out. That's why devices need the right batteries.

Batteries are labeled with a nominal voltage. OpenStax Chemistry gives about 1.5 V for a zinc and manganese dioxide dry cell of any size (D, C, AA, AAA): larger cells hold more reactants and can move more charge, not push harder. A real battery's terminal voltage also falls as it is depleted, because its internal resistance rises. OpenStax Physics notes that cells are commonly connected in series in flashlights, toys and other appliances to produce a larger total emf.

OpenStax also gives the real-world limit the model ignores: a 25 W bulb designed for 120 V, connected to 240 V, briefly glows very brightly and burns out. Doubling its voltage nearly quadruples its power to about 100 W; it would be exactly 100 W if the resistance stayed constant, but the hotter filament resists more.