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A resistor \(R_{1}\) consumes electrical power \(P_{1}\) when con nected to an emf \(\mathcal{E}\). When resistor \(R_{2}\) is connected to the same emf, it consumes electrical power \(P_{2}\). In terms of \(P_{1}\) and \(P_{2}\), what is the total electrical power consumed when they are both connected to this emf source (a) in parallel and (b) in series?

Short Answer

Expert verified
In parallel, total power is \( P_1 + P_2 \). In series, total power is \( \frac{P_1 \cdot P_2}{P_1 + P_2} \).

Step by step solution

01

Understanding Power in Resistors

The power consumed by each resistor connected to an emf source \( \mathcal{E} \) can be given by the formula \( P = \frac{\mathcal{E}^2}{R} \). For resistor \( R_1 \), the power is \( P_1 = \frac{\mathcal{E}^2}{R_1} \), which gives us \( R_1 = \frac{\mathcal{E}^2}{P_1} \). Similarly, for resistor \( R_2 \), \( R_2 = \frac{\mathcal{E}^2}{P_2} \).
02

Calculate Total Power in Parallel

When resistors are connected in parallel, the total power consumed is the sum of individual powers. The formula used in this arrangement is that the total power \( P_{\text{parallel}} = P_1 + P_2 \), since each resistor sees the same voltage \( \mathcal{E} \).
03

Calculate Total Resistance in Series

For series circuits, the total resistance is the sum of individual resistances: \( R_{\text{series}} = R_1 + R_2 = \frac{\mathcal{E}^2}{P_1} + \frac{\mathcal{E}^2}{P_2} \). To find this in terms of \( P_1 \) and \( P_2 \), simplify later.
04

Total Power in Series

Using the formula \( P_{\text{series}} = \frac{\mathcal{E}^2}{R_{\text{series}}} \), substitute the value of \( R_{\text{series}} \) from Step 3: \( P_{\text{series}} = \left( \frac{P_1 \cdot P_2}{P_1 + P_2} \right) \).
05

Express Total Power in Terms of P1 and P2

For the series circuit, we simplify the expression \( P_{\text{series}} = \left( \frac{P_1 \cdot P_2}{P_1 + P_2} \right) \) to be in terms of \( P_1 \) and \( P_2 \) directly, as this finds the total power dissipated when both resistors are in series.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parallel Circuits and Electrical Power
In a parallel circuit configuration, multiple resistors are connected across the same voltage source. The defining characteristic of parallel circuits is that each component sees the same voltage across it, similar to attaching several appliances to a single power outlet in your home.
This arrangement affects electric power, which can be understood by looking at each component separately. In our problem, when the resistors are connected in parallel to the electromotive force (emf) \(\mathcal{E}\), each resistor operates independently with the same voltage across it. Thus, the total power consumed is simply the sum of the powers consumed by each resistor.
  • Each resistor encounters voltage \(\mathcal{E}\) across it.
  • The formula for power for each resistor is given by \(P = \frac{\mathcal{E}^2}{R}\).
  • Total power in parallel: \(P_{\text{parallel}} = P_1 + P_2\).
If resistor \(R_1\) consumes power \(P_1\) and resistor \(R_2\) consumes power \(P_2\), the total power is simply \(P_1 + P_2\). This neat addition of individual powers highlights how each path in a parallel setup acts separately yet contributes to the overall circuit power.
Series Circuits and Electrical Power
In a series circuit, resistors are connected end-to-end, forming a single path for the electric current to flow. The key feature here is that the same current flows through all components in the series, but the voltage across each component can vary.
For our given problem, when resistors \(R_1\) and \(R_2\) are tipped in series with the emf \(\mathcal{E}\), the entire voltage is divided across the resistors depending on their resistance values. Unlike in parallel circuits where voltages are independent, the series setup results in the individual voltages adding up to the source voltage.
  • Total resistance: \(R_{\text{series}} = R_1 + R_2\).
  • The same current flows through both resistors.
  • Total power in series: \(P_{\text{series}} = \left(\frac{P_1 \cdot P_2}{P_1 + P_2}\right)\).
Instead of adding powers directly, we use combined resistances to find the power consumed. This is because the voltage drop is shared, affecting the power seen across the resistors. Using the relation between power, resistance, and voltage, the formula \(P_{\text{series}} = \frac{P_1 \cdot P_2}{P_1 + P_2}\) gives us the effective power consumption when resistors are serially coupled, demonstrating different principles than parallel combinations.
Understanding Resistive Circuits
A resistive circuit is one where the entire electrical behavior can be defined by resistors. These circuits offer a fundamental way to understand how power dissipates in various configurations, including both parallel and series setups.
Resistors limit the flow of current, converting electrical energy into thermal energy, which can quickly explain why they might feel warm to the touch. When dealing with resistive circuits, understanding Ohm's Law and the power formula are foundational, providing insight into how resistors affect voltage, current, and power in a circuit.
  • Ohm's Law: \(V = IR\), relates the voltage across a resistor to the current through it and its resistance.
  • Power formula: \(P = VI = I^2 R = \frac{V^2}{R}\), shows how power is affected by current, voltage, and resistance.
  • In resistive circuits, energy is lost to heat as current flows through resistance.
By applying these principles, we can make calculations and predictions about how circuits will behave. It's essential to grasp these fundamentals, as they create the groundwork for more complex electrical engineering concepts, making resistive circuits a crucial part of learning about electricity and power.

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Most popular questions from this chapter

A car battery with a \(12 \mathrm{~V}\) emf and an internal resistance of \(0.04 \Omega\) is being charged wtih a current of \(50 \mathrm{~A}\) (a) Find the potential difference \(V\) across the terminals of the battery. (b) Find the rate at which energy is being dissipated as heat inside the battery. (c) Find the rate of energy conversion from electrical to chemical.

Light Bulbs in Series. A 60-W, \(120-\mathrm{V}\) light bulb and a \(200-\mathrm{W}, 120-\mathrm{V}\) light bulb are connected in series across a \(240-\mathrm{V}\) line. Assume that the resistance of each bulb does not vary with current. (Note: This description of a light bulb gives the power it dissipates when connected to the stated potential difference; that is, a \(25-\mathrm{W}, 120-\mathrm{V}\) light bulb dissipates \(25 \mathrm{~W}\) when connected to a 120-V line.) (a) Find the current through the bulbs. (b) Find the power dissipated in each bulb. (c) One bulb burns out very quickly. Which one? Why?

An \(R-C\) circuit has a time constant \(R C\). (a) If the circuit is discharging, how long will it take for its stored energy to be reduced to \(1 / e\) of its initial value? (b) If it is charging, how long will it take for the stored energy to reach \(1 / e\) of its maximum value?

A battery has an emf of \(\mathcal{E}=6 \mathrm{~V}\). The battery is connected in series with an ammeter and a voltmeter. If a certain resistor is connected in parallel with the voltmeter, the voltmeter reading decreases by a factor of 3, and the ammeter reading increases by the same factor. What is the initial reading \(V\) of the voltmeter? All elements of the circuit have unknown internal resistances.

A \(1500-W\) electric heater is plugged into the outlet of a \(120-\mathrm{V}\) circuit that has a \(20-\mathrm{A}\) circuit breaker. You plug an electric hair dryer into the same outlet. The hair dryer has power settings of \(600 \mathrm{~W}, 900 \mathrm{~W}, 1200 \mathrm{~W}\), and \(1500 \mathrm{~W}\). You start with the hair dryer on the \(600-\mathrm{W}\) setting and increase the power setting until the circuit breaker trips. What power setting caused the breaker to trip?

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