Elementary circuits

Explain in your own words the difference between connecting two light bulbs in series and connecting them in parallel, and give an everyday example of each type of connection.

A light bulb is connected to a $4,5\ \text{V}$ battery and a current of $0,3\ \text{A}$ flows through its filament. Applying Ohm's law, calculate the resistance of the filament.

Two resistors of $100\ \Omega$ and $200\ \Omega$ are connected in series to a $12\ \text{V}$ battery. Calculate the total resistance of the circuit and the current flowing through it.

A circuit has a $9\ \text{V}$ battery and a single resistor of $100\ \Omega$. Applying Ohm's law, $V = I \cdot R$, calculate the current flowing through it.

A current of $0,5\ \text{A}$ flows through a resistor of $20\ \Omega$. Calculate the voltage across its terminals.

Two resistors of $100\ \Omega$ and $300\ \Omega$ are connected in series to a $12\ \text{V}$ battery. Calculate the current flowing through the circuit and the voltage across the terminals of each resistor, verifying that the two voltages add up to the battery voltage.

Two identical light bulbs are connected in parallel to a $4,5\ \text{V}$ battery, and a current of $0,2\ \text{A}$ flows through each one. What total current does the battery supply? What is the resistance of each light bulb?

Consider two setups: the first consists of two resistors in series ($R_1 = 220\ \Omega$ and $R_2 = 330\ \Omega$) connected to a $9\text{ V}$ battery with an ammeter; the second consists of two light bulbs in parallel connected to a $4,5\text{ V}$ battery. Calculate the current shown on the ammeter in the series setup and explain what happens in the parallel setup if one of the light bulbs burns out.

Two identical resistors are connected in series to a $6\ \text{V}$ battery. Without knowing their value, reason what voltage a voltmeter would measure across the terminals of each one and why.

A parallel circuit powered at $10\ \text{V}$ has two branches: one with a resistor of $50\ \Omega$ and another with one of $100\ \Omega$. Calculate the current in each branch and the total current, and reason why the branch with less resistance carries more current.

You want to protect an LED diode ($V_F = 2\ \text{V}$, nominal operating current $I = 20\ \text{mA}$, and maximum destructive current of $30\ \text{mA}$) connected to a $9\ \text{V}$ battery. In the workshop, you have $1/4\ \text{W}$ ($0,25\ \text{W}$) resistors from the standard commercial series: $220\ \Omega$, $330\ \Omega$, $390\ \Omega$, and $470\ \Omega$.

1. a) Calculate the exact theoretical value of the required resistor and justify which of the available commercial values you should choose to ensure that the LED operates properly without being destroyed.
2. b) With the chosen commercial value, calculate the actual current that will flow through the circuit and check whether the $0,25\ \text{W}$ resistor can handle the power it must dissipate without burning out.
3. c) Predict what would happen to the LED if the $220\ \Omega$ resistor were used by mistake.

You have a $9\ \text{V}$ battery and three identical incandescent light bulbs whose nominal ratings are $9\ \text{V}$ and $4,5\ \text{W}$ ($18\ \Omega$ resistance each). A student states: "Since the light bulbs share the effort in series, the battery will last exactly the same in series as in parallel, even though they shine brighter in parallel."

1. a) Quantitatively prove whether the statement about the battery's lifespan is true or false by calculating the total power supplied by the battery in both setups.
2. b) Calculate the power dissipated by each bulb in the series setup and in the parallel setup, and deduce from your results why and in what proportion they shine brighter in one of the two cases.
3. c) Predict and quantitatively justify what will happen to the total battery current and to the brightness of the remaining bulbs if one bulb is unscrewed in the series circuit, and if the same is done in the parallel circuit.

A mixed circuit consists of a bulb A connected in series with a parallel combination formed by two other bulbs, B and C. The three bulbs are identical ($R = 20\ \Omega$ each), and the circuit is powered by a $12\ \text{V}$ power source.

1. a) Calculate the current flowing through each bulb and the voltage across the terminals of each of them.
2. b) A classmate claims: "Bulb A consumes part of the battery's current, which is why only half the current reaches the branches of B and C." Refute this statement scientifically by explaining how current behaves in the circuit according to the principle of conservation of charge.
3. c) Predict with reasoning what will happen to the brightness of bulb A and that of bulb B if bulb C suddenly burns out (open circuit in that branch). Justify your answer by calculating the new current values.

Analyse the mixed circuit in the figure:

1. a) Calculate the total current $I$ supplied by the battery and the voltage across the parallel combination ($R_2$ and $R_3$).
2. b) Quantitatively predict what would happen to the current supplied by the battery and to the power dissipated by $R_1$ if an accidental short circuit occurs across the terminals of $R_3$ (a zero-resistance bridge between its terminals). Explain what danger this poses to $R_1$.
3. c) What resistance value should a new component $R_4$, placed in parallel with $R_2$ and $R_3$, have so that the total current $I$ of the circuit increases to exactly $0,75\ \text{A}$?

Two incandescent light bulbs with resistances of $10\ \Omega$ and $40\ \Omega$ are connected to a $12\ \text{V}$ power source. A student claims: "The $10\ \Omega$ bulb will always shine brighter than the $40\ \Omega$ one, whether we connect them in series or in parallel, because it offers less resistance and lets more energy through."

1. a) Calculate the power dissipated by each bulb when connected in series and determine which one shines brighter.
2. b) Calculate the power dissipated by each bulb when connected in parallel and determine which one shines brighter.
3. c) Critically analyse the student's reasoning: explain why they are right in one setup but mistaken in the other, justifying which power formula ($P = I^2 \cdot R$ or $P = V^2 / R$) is convenient to use in each case to understand the result.

In the workshop, a $12\ \text{V}$ lamp with a resistance of $24\ \Omega$ is connected to a $12\ \text{V}$ power supply. You have a digital multimeter with an internal $2\ \text{A}$ protective fuse and an internal resistance as an ammeter of $0,1\ \Omega$.

1. a) If a student mistakenly connects the ammeter in parallel with the bulb instead of in series, calculate the current that will pass through the bulb and the current that will flow through the ammeter.
2. b) Based on your calculations, predict and justify whether the bulb will light up and what will happen to the measuring device.
3. c) Describe or draw the standard schematic to simultaneously measure the voltage across the bulb (with a voltmeter) and the current it consumes (with an ammeter), justifying what ideal value the internal resistance of each instrument should have.