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Given a 48.0-V battery and 24.0-Ω and 96.0-Ω resistors, find the current and power for each when connected in series. (b) Repeat when the resistances are in parallel. Please show all work using this. (Actual values, equations, and drawings)

Given a 48.0-V battery and 24.0-Ω and 96.0-Ω resistors, find the current and power for each when connected in series. (b) Repeat when the resistances are in parallel. Please show all work using this. (Actual values, equations, and drawings)

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Publisher:Raymond A. Serway, Chris Vuille
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30)(a) Given a 48.0-V battery and 24.0-Ω and 96.0-Ω resistors, find the current and power for each when connected in series. (b) Repeat when the resistances are in parallel.

Please show all work using this. (Actual values, equations, and drawings)

 

The image contains handwritten notes on electrical circuits, specifically on series and parallel resistor configurations. Below is the transcribed content, along with explanations of the diagrams described in the text. --- **Problem Statement:** \[ E_b = 48V \] \[ R_1 = 24\Omega \] \[ R_2 = 96\Omega \] **Tasks:** 1. Calculate \( I_s \) and \( J_s \) for the series circuit. 2. Calculate \( I_p \) and \( J_p \) for the parallel circuit. --- **a) Series Circuit Analysis:** - **Diagram:** A series circuit with a voltage source (\( E_b \)) and two resistors (\( R_1 \) and \( R_2 \)) in series. - **Equations:** \[ R_{eqs} = R_1 + R_2 \] \[ I_s = \frac{E_b}{R_{eqs}} = \frac{E_b}{R_1 + R_2} \] \[ J_s = E_b \cdot I_s = \frac{E_b^2}{R_1 + R_2} \] --- **b) Parallel Circuit Analysis:** - **Diagram:** A parallel circuit with a voltage source (\( E_b \)) and two resistors (\( R_1 \) and \( R_2 \)) parallel to each other. - **Equations:** \[ \frac{1}{R_{eqp}} = \frac{1}{R_1} + \frac{1}{R_2} \] \[ R_{eqp} = \frac{R_1 \cdot R_2}{R_1 + R_2} \] \[ I_p = \frac{E_b}{R_{eqp}} = \frac{E_b \cdot (R_1 + R_2)}{R_1 \cdot R_2} \] \[ J_p = E_b \cdot I_p = \frac{E_b^2 \cdot (R_1 + R_2)}{R_1 \cdot R_2} \] --- These notes provide a clear comparison between the behavior of series and parallel circuits, specifically in terms of equivalent resistance, current, and power calculations.
Transcribed Image Text:The image contains handwritten notes on electrical circuits, specifically on series and parallel resistor configurations. Below is the transcribed content, along with explanations of the diagrams described in the text. --- **Problem Statement:** \[ E_b = 48V \] \[ R_1 = 24\Omega \] \[ R_2 = 96\Omega \] **Tasks:** 1. Calculate \( I_s \) and \( J_s \) for the series circuit. 2. Calculate \( I_p \) and \( J_p \) for the parallel circuit. --- **a) Series Circuit Analysis:** - **Diagram:** A series circuit with a voltage source (\( E_b \)) and two resistors (\( R_1 \) and \( R_2 \)) in series. - **Equations:** \[ R_{eqs} = R_1 + R_2 \] \[ I_s = \frac{E_b}{R_{eqs}} = \frac{E_b}{R_1 + R_2} \] \[ J_s = E_b \cdot I_s = \frac{E_b^2}{R_1 + R_2} \] --- **b) Parallel Circuit Analysis:** - **Diagram:** A parallel circuit with a voltage source (\( E_b \)) and two resistors (\( R_1 \) and \( R_2 \)) parallel to each other. - **Equations:** \[ \frac{1}{R_{eqp}} = \frac{1}{R_1} + \frac{1}{R_2} \] \[ R_{eqp} = \frac{R_1 \cdot R_2}{R_1 + R_2} \] \[ I_p = \frac{E_b}{R_{eqp}} = \frac{E_b \cdot (R_1 + R_2)}{R_1 \cdot R_2} \] \[ J_p = E_b \cdot I_p = \frac{E_b^2 \cdot (R_1 + R_2)}{R_1 \cdot R_2} \] --- These notes provide a clear comparison between the behavior of series and parallel circuits, specifically in terms of equivalent resistance, current, and power calculations.
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