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A resistor \(R_{1}\) consumes electrical power \(P_{1}\) when connected 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
(a) Parallel: \(P_{1} + P_{2} \); (b) Series: \(\frac{P_{1} P_{2}}{P_{1} + P_{2}}\).

Step by step solution

01

Understanding Power in Parallel

When two resistors are connected in parallel to the same emf \( \mathcal{E} \), they both receive full voltage from the emf. The power consumed by each resistor remains the same as it would individually when connected to the emf source. Thus, the total power \( P_{\text{total, parallel}} \) consumed is simply the sum of the individual powers: \[ P_{\text{total, parallel}} = P_{1} + P_{2}. \]
02

Equation for Power in Series

For resistors in series, the emf \( \mathcal{E} \) is distributed across the resistors based on their resistance, not directly across each one as in parallel. Each resistor experiences less voltage, but because the total current is the same through both, we analyze the equivalent resistance to find total power.The combined resistance is \( R_{\text{total}} = R_{1} + R_{2} \). The current through this network is \( I = \frac{\mathcal{E}}{R_{\text{total}}} \). Hence, the total power consumed is:\[ P_{\text{total, series}} = I^2 \times R_{\text{total}} = \left(\frac{\mathcal{E}}{R_{\text{total}}}\right)^2 \times (R_{1} + R_{2}). \]
03

Expressing Series Power in Known Terms

Since the power given for each resistor when alone are \( P_{1} = \frac{\mathcal{E}^2}{R_{1}} \) and \( P_{2} = \frac{\mathcal{E}^2}{R_{2}} \), we can solve for \( R_{1} \) and \( R_{2} \) to find expressions:\[ R_{1} = \frac{\mathcal{E}^2}{P_{1}} \] \[ R_{2} = \frac{\mathcal{E}^2}{P_{2}}. \]Substitute these into the series power equation:\[ P_{\text{total, series}} = \frac{\mathcal{E}^2}{\left(\frac{\mathcal{E}^2}{P_{1}} + \frac{\mathcal{E}^2}{P_{2}}\right)} \left( \frac{\mathcal{E}^2}{P_{1}} + \frac{\mathcal{E}^2}{P_{2}} \right) = \frac{P_{1} P_{2}}{P_{1} + P_{2}}. \]

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Key Concepts

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

Power in Parallel Circuits
When analyzing power in parallel circuits, it is crucial to understand how electrical components share an emf (or voltage source). In a parallel arrangement, each resistor is directly connected across the voltage source. This means
  • Each resistor experiences the full voltage of the source.
  • Resistors do not affect each other's voltage, as would happen in series.
The overall power consumed in a parallel circuit is calculated simply by summing the power consumed by each individual component:\[P_{\text{total, parallel}} = P_1 + P_2.\]In this equation:
  • \(P_1\) is the power consumed by the first resistor \(R_1\).
  • \(P_2\) is the power consumed by the second resistor \(R_2\).
Hence, a key advantage of parallel circuits is that each component receives the same voltage, making it straightforward to determine the total power consumed.
Power in Series Circuits
Series circuits behave differently than parallel circuits. In a series configuration, each resistor does not experience the full voltage across the emf. Instead, the total voltage is divided among the components according to their resistances:
  • The current flowing through each resistor is the same.
  • The voltage drop across each resistor can vary.
The total equivalent resistance in a series circuit is the sum of all individual resistances:\[R_{\text{total}} = R_1 + R_2.\]With this, you can calculate the total current using Ohm's law:\[I = \frac{\mathcal{E}}{R_{\text{total}}}.\]Substituting this current into the power formula gives us the expression for the total power consumption:\[P_{\text{total, series}} = I^2 \times R_{\text{total}},\]which can be reduced to:\[P_{\text{total, series}} = \frac{P_1 P_2}{P_1 + P_2},\]by expressing resistances in terms of their individual power values. This equation shows that, unlike in parallel circuits, power in a series circuit is determined by the individual resistance values relative to each other.
Equivalent Resistance
The concept of equivalent resistance is vital in circuit analysis, as it simplifies complex circuits into a single resistor that has the same effect on the circuit as all the resistors combined. This is beneficial for both series and parallel circuits:
**For Series Circuits:**
  • Equivalent resistance is the sum of all resistances: \( R_{\text{total}} = R_1 + R_2 + ... + R_n \).
  • This leads to a straightforward calculation of total current using Ohm's law.
**For Parallel Circuits:**
  • The inverse of the total resistance is the sum of the inverses of each individual resistance:\[\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n}.\]
  • This approach confirms that equivalent resistance in parallel is always less than the smallest individual resistance.
Understanding equivalent resistance is essential for predicting how a circuit will behave in terms of both voltage distributions and current flow. It also helps in calculating the overall power consumed.

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Most popular questions from this chapter

A. \(2.36-\mu F\) capacitor that is initially uncharged is connected in series with a \(4.26-\Omega\) resistor and an emf source with \(\mathcal{E}=120 \mathrm{V}\) and negligible intermal resistance. (a) Just after the connection is made, what are (i) the rate at which electrical energy is being dissipated in the resistor; (ii) the rate at which the electrical energy stored in the capacitor is increasing; (iii) the electrical energy stored in the capacitor is increasing; (iii) the electrical power output of the source? How do the answers to parts (i), (ii), and (iii) compare? (b) Answer the same questions as in part (a) at a long time after the connection is made. (c) Answer the same questions as in part (a) at the instant when the charge on the capacitor is one-half its final value.

Three identical resistors are connected in series. When a certain potential difference is applied across the combination, the total power dissipated is 27 \(\mathrm{W}\) . What power would be dissipated if the three resistors were connected in parallel across the same potential difference?

A \(1.50-\mu F\) capacitor is charging through a \(12.0-\Omega\) resistor using a \(10.0-\mathrm{V}\) battery. What will be the current when the capacitor has acquired \(\frac{1}{4}\) of its maximum charge? Will it be \(\frac{1}{4}\) of the maximum current?

An emf source with \(\mathcal{E}=120 \mathrm{V},\) a resistor with \(R=80.0 \Omega\) and a capacitor with \(C=4.00 \mu F\) are connected in series. As the capacitor charges, when the current in the resistor is 0.900 \(\mathrm{A}\) , what is the magnitude of the charge on each plate of the capacitor?

A capacitor is charged to a potential of 12.0 \(\mathrm{V}\) and is then connected to a voltmeter having an internal resistance of 3.40 \(\mathrm{M} \Omega\) . After a time of 4.00 s the voltmeter reads 3.0 \(\mathrm{V}\) . What are (a) the capacitance and \((\mathrm{b})\) the time constant of the circuit?

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