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Two resistors, 42.0 and 64.0 \Omega, are connected in parallel. The current through the \(64.0-\Omega\) resistor is 3.00 A. (a) Determine the current in the other resistor. (b) What is the total power supplied to the two resistors?

Short Answer

Expert verified
(a) The current in the 42.0 \(\Omega\) resistor is approximately 4.57 A. (b) The total power supplied is approximately 1452 W.

Step by step solution

01

Understanding Ohm's Law and Parallel Circuits

In a parallel circuit, the voltage across each component is the same. Ohm's Law states that \(V = I \times R\), where \(V\) is voltage, \(I\) is current, and \(R\) is resistance. Given the current through the \(64.0\, \Omega\) resistor is \(3.00\, A\), use Ohm's Law to find the voltage across this resistor.
02

Calculate Voltage Across the 64.0 \(\Omega\) Resistor

Using Ohm’s Law \(V = I \times R\) for the \(64.0 \, \Omega\) resistor:\[ V = 3.00 \, A \times 64.0 \, \Omega = 192 \, V \]
03

Determine Current Through the 42.0 \(\Omega\) Resistor

Since the voltage across each resistor in a parallel circuit is the same, the voltage across the \(42.0 \, \Omega\) resistor is also \(192 \, V\). Using Ohm's Law \(V = I \times R\) again to find the current:\[ I = \frac{V}{R} = \frac{192 \, V}{42.0 \, \Omega} \approx 4.57 \, A \]
04

Calculate Total Power Supplied

The total power supplied to parallel resistors can be calculated using the formula:\[ P = V \times I_{\text{total}} \]First, find the total current \(I_{\text{total}}\) by summing the currents through each resistor:\[ I_{\text{total}} = 3.00 \, A + 4.57 \, A = 7.57 \, A \]Then calculate the total power:\[ P = 192 \, V \times 7.57 \, A \approx 1452 \, W \]

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Key Concepts

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

Parallel Circuits
When dealing with parallel circuits, it's important to understand that each component within the circuit is connected across the same pair of nodes, meaning that all components experience the same voltage. In parallel circuits, the total current flowing through the circuit is divided among the different paths. Each path can have different resistances, which will influence how the total current is distributed among the paths.
A few key properties of parallel circuits include:
  • The voltage across all components in a parallel circuit is identical.
  • The total current is the sum of the currents through each parallel component.
  • The equivalent or total resistance (\( R_{ ext{total}} \)) in parallel is always less than the smallest resistance in the circuit and can be calculated using the formula: \[\frac{1}{R_{ ext{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots + \frac{1}{R_n}\]
Understanding these concepts is crucial when working with Ohm's Law within parallel circuits, as it ensures accurate calculations of current, voltage, and resistance.
Resistor Current Calculation
Calculating the current through a resistor in a parallel circuit starts with applying Ohm's Law to the known values. Ohm's Law, given by \( V = I \times R \), allows you to find missing variables by rearranging the equation based on the values available.
For example, to find the current through a resistor with a known resistance and voltage, you rearrange the formula to \( I = \frac{V}{R} \). This relationship highlights how currents in parallel circuits are affected by the resistance of each path. A larger resistance leads to a smaller current, while smaller resistance allows for a larger current, given constant voltage across them.Here's a step-by-step approach:
  • Find the voltage across the resistors in the circuit, which is consistent across parallel branches.
  • Use the voltage and the resistance of each resistor to calculate the current using \( I = \frac{V}{R} \).
  • Apply these calculations to each component to determine individual currents.
Ensuring these calculations are correct will provide the foundation for further analyses in the circuit.
Power in Electrical Circuits
Power in electrical circuits, especially those containing parallel circuits, is another crucial concept. Power (\( P \)) in a circuit is defined as the rate at which energy is used or converted and can be calculated using \[ P = V \times I \]. Understanding how to compute power helps in determining the efficiency and safety of electrical components.In a parallel circuit:
  • The power across each resistor can be calculated individually by applying the formula \( P = V \times I \), where voltage (\( V \)) is the same across all resistors.
  • The total power supplied to the circuit is the sum of the powers consumed by each resistor. This is critical in evaluating the total energy expenditure in the electrical system.
  • Be mindful of different power values represented in units like watts, kilowatts, etc.
Keeping track of power distribution is essential for ensuring that circuits are not overloaded, which is critical in both household and industrial applications.

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Most popular questions from this chapter

The recovery time of a hot water heater is the time required to heat all the water in the unit to the desired temperature. Suppose that a 52 -gal \(\left(1.00 \mathrm{gal}=3.79 \times 10^{-3} \mathrm{m}^{3}\right)\) unit starts with cold water at \(11^{\circ} \mathrm{C}\) and delivers hot water at \(53{ }^{\circ} \mathrm{C}\). The unit is electric and utilizes a resistance heater \((120 \mathrm{V}\) ac, \(3.0 \Omega)\) to heat the water. Assuming that no heat is lost to the environment, determine the recovery time (in hours) of the unit.

Three resistors, \(25,45,\) and \(75 \Omega,\) are connected in series, and a \(0.51-\) A current passes through them. What are (a) the equivalent resistance and (b) the potential difference across the three resistors?

Two capacitors are connected to a battery. The battery voltage is \(V=60.0 \mathrm{V},\) and the capacitances are \(C_{1}=2.00 \mu \mathrm{F}\) and \(C_{2}=4.00 \mu \mathrm{F}\) Determine the total energy stored by the two capacitors when they are wired (a) in parallel and (b) in series.

The rms current in a copy machine is \(6.50 \mathrm{A}\), and the resistance of the machine is \(18.6 \Omega .\) What are (a) the average power and (b) the peak power delivered to the machine?

Three capacitors are connected in series. The equivalent capacitance of this combination is \(3.00 \mu \mathrm{F}\). Two of the individual capacitances are \(6.00 \mu \mathrm{F}\) and \(9.00 \mu \mathrm{F} .\) What is the third capacitance (in \(\mu \mathrm{F}) ?\)

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