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Q38PE

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Found in: Page 776

College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

Find the currents flowing in the circuit in Figure$$21.52$$. Explicitly show how you follow the steps in the Problem-Solving Strategies for Series and Parallel Resistors.

The currents flowing in the circuit

\begin{align}{}{I_1} = - 0.321\;A\\{I_2} = 0.248\;A\\{I_3} = - 0.073\;A\end{align}

See the step by step solution

Step 1: Concept Introduction

In a series circuit, the current in each resistor is the same because the output

current of the first resistor flows into the input of the second resistor.

In a parallel circuit, all of the resistor leads on one side are linked together,

and all of the resistor leads on the other side are linked together.

Step 2: Given information

\begin{align}{}{E_1} = 24\;\;V\\{R_1} = 5\;\Omega \\{r_1} = 0.10\;\Omega \\{E_4} = 36\;\;V\end{align}

\begin{align}{}{E_2} = 48\;\;V\\{R_2} = 40\;\Omega \\{r_2} = 0.50\;\Omega \\{R_5} = 20\;\Omega \end{align}

\begin{align}{}{E_3} = 6\;\;V\\{R_3} = 78\;\Omega \\{r_3} = 0.50\;\Omega \\{r_4} = 0.20\;\Omega \end{align}

Step 3: Explanation

In the figure $$21.52$$an equation for loop abcdefghija may be found using the loop rule:

$$- {I_1}{R_1} + {E_1} - {r_1}{I_1} - {R_5}{I_1} - {r_4}{I_3} - {E_4} - {r_3}{I_3} + {E_3} - {R_3}{I_3} = 0$$

Sources acting in the same direction as the current flow are proclaimed positive, whereas resistances and sources acting in the opposite direction are declared negative.

$$- {I_1} \times (5 + 0.1 + 20) - {I_3} \times (0.2 + 0.05 + 78) + 24 - 36 + 6 = 0$$

For loop aklefghija, the following equation is created:

$$- {I_2} \times (0.5 + 40) - {I_3} \times (0.2 + 0.05 + 78) + 48 - 36 + 6 = 0$$

And final equation:

\begin{align}{ }{I_3} = {I_1} + {I_2}\\{I_1} \times 65.6 - {I_3}40.5 = 24\end{align}

When we plug this into the first equation, we get:

$${I_3} = - 0.073\;A$$

To get $${{\rm{I}}_{\rm{1}}}$$, use that solution:

$${I_1} = - 0.321\;A$$

Finally:

$${I_2} = 0.248\;A$$

Therefore,the currents values are:

\begin{align}{}{I_1} = - 0.321\;A\\{I_2} = 0.248\;A\\{I_3} = - 0.073\;A\end{align}