Series and parallel circuits

Two light bulbs are connected in series, one after the other, to a battery, as shown in the figure.

Through which of the two does more current flow? Justify your answer.

Two resistors of $2\ \Omega$ and $3\ \Omega$ are connected in series, as shown in the figure.

What is the total (equivalent) resistance of the combination?

In which type of connection does current have only one possible path, series or parallel? And in which can it be split among several branches?

A simple flashlight has a battery, a switch and a light bulb.

1. a) Are they connected in series or in parallel?
2. b) What happens to the current when we open the switch?

Two resistors, $R_1 = 2\ \Omega$ and $R_2 = 4\ \Omega$, are connected in series to a $12\ \text{V}$ battery, as shown in the figure.

1. a) Calculate the equivalent resistance.
2. b) Calculate the current flowing through the circuit.
3. c) Find the voltage across each resistor using Ohm's law and verify that the two voltages add up to the $12\ \text{V}$ of the battery.

Three light bulbs of $2\ \Omega$ each are connected in series to a $12\ \text{V}$ battery, as shown in the figure.

1. a) Calculate the total resistance.
2. b) What current passes through them?
3. c) What voltage is across each light bulb?

Two identical light bulbs in series are connected to a $9\ \text{V}$ battery.

1. a) Without knowing their resistance, reason how the voltage is shared between them.
2. b) How many volts are across each one?
3. c) How would the sharing change if the bulbs were different?

We connect identical $3\ \Omega$ light bulbs in series to a $9\ \text{V}$ battery, adding one at a time.

1. a) Calculate the circuit current with one, with two, and with three light bulbs.
2. b) What happens to the brightness of the bulbs as we add more in series?
3. c) Explain the result using equivalent resistance and Ohm's law.

A classmate states: "In series, the current from the battery is shared between the bulbs."

1. a) Is he right?
2. b) With a $12\ \text{V}$ battery and two $3\ \Omega$ light bulbs in series, calculate the current passing through each one and the voltage across each one.
3. c) Correct your classmate's statement: what is actually shared in a series circuit?

Two light bulbs are connected in parallel to the same battery, each along its own path, as shown in the figure.

What do they have in common electrically: the current passing through them or the voltage across their ends? Explain.

Two resistors, $R_1 = 6\ \Omega$ and $R_2 = 12\ \Omega$, are connected in parallel to a $12\ \text{V}$ battery.

1. a) What voltage is across the ends of each one?
2. b) Calculate the current flowing through each branch.
3. c) What is the total current supplied by the battery?

Two resistors of $4\ \Omega$ and $12\ \Omega$ are in parallel. Knowing that in this setup $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$ holds, calculate the equivalent resistance of the combination.

We want to measure the total current supplied by a battery in a circuit with two light bulbs in parallel, as shown in the figure.

1. a) Where should we place the ammeter: in one of the branches or next to the battery, before the path divides?
2. b) What would the device measure in each of those two positions?

You have a $6\ \text{V}$ battery and two identical $6\ \Omega$ light bulbs, as shown in the figure.

1. a) If you connect them in series, what current flows through each one?
2. b) If you connect them in parallel, what current flows through each one?
3. c) In which setup do they shine brighter? Justify your answer using the results.

Now we connect identical $12\ \Omega$ light bulbs in parallel to a $6\ \text{V}$ battery, adding one at a time.

1. a) What current passes through each bulb when there are one, two, or three?
2. b) Calculate the total current supplied by the battery in each case.
3. c) Explain why the battery runs out faster the more bulbs we connect in parallel.

A resistor of $6\ \Omega$ and another of $3\ \Omega$ are connected in parallel.

1. a) Calculate the equivalent resistance.
2. b) Verify that it is even smaller than the smaller of the two.
3. c) Explain, thinking about the paths available for the current, why adding a branch in parallel always makes it easier for current to flow.

In an old set of string lights, the bulbs are connected in series, as shown in the figure.

If one bulb burns out and its filament breaks, what happens to the others? Explain by thinking about the path followed by the current.

In a scale model, two light bulbs are connected in parallel to a battery. If you unscrew one of them, what happens to the other? Justify your answer.

In the classroom there are six LED tubes, and when one breaks down, the others stay on.

1. a) Are they connected in series or in parallel?
2. b) Justify your answer by explaining what happens to the current when one fails.

In homes, outlets and lamps are connected in parallel and not in series. Give a reason, thinking about what would happen when switching off an appliance or if one breaks down.

The two headlights of a car are connected in parallel to the $12\ \text{V}$ battery, and each one has a resistance of $6\ \Omega$.

1. a) What current flows through each headlight?
2. b) What is the total current supplied by the battery with both headlights on?
3. c) If one headlight burns out, what happens to the other? Justify your answer considering the type of connection.

You are going to light up two scale models with one bulb each, and they must be able to be switched on and off separately using a single battery.

1. a) Would you connect them in series or in parallel? Why?
2. b) What advantage does your choice have if a bulb burns out?
3. c) What disadvantage does it pose for the battery life? Relate this to the total current supplied.

In a home, the oven ($23\ \Omega$) and the microwave ($46\ \Omega$) are connected in parallel to the $230\ \text{V}$ mains.

1. a) Calculate the current flowing through each appliance when they operate simultaneously.
2. b) What total current enters through the kitchen circuit?
3. c) Explain why it is convenient for them to be in parallel rather than in series, considering the voltage that each appliance requires.

Three resistors of $2\ \Omega$, $4\ \Omega$ and $4\ \Omega$ are in parallel, as shown in the figure.

1. a) Calculate the equivalent resistance using $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$.
2. b) If the combination is connected to a $6\ \text{V}$ battery, what total current flows?
3. c) Calculate the current in each branch and check that the three sum to the total current.