Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN: 9780133923605
Author: Robert L. Boylestad
Publisher: PEARSON
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![**8. Determine the value of the current through R2 in the circuit shown.**
Given:
- \( I_T = 12 \, \text{mA} \)
- \( I_1 = 3 \, \text{mA} \)
- \( I_3 = 2 \, \text{mA} \)
**Diagram Explanation:**
The diagram illustrates a parallel circuit with three resistors: R1, R2, and R3. Current \( I_T \) enters the circuit and splits into three separate currents denoted as \( I_1 \), \( I_2 \), and \( I_3 \), flowing through R1, R2, and R3 respectively. The main current source \( I_T \) is given as 12 mA, with \( I_1 \) as 3 mA, and \( I_3 \) as 2 mA.
To determine the current through R2 (\( I_2 \)), use Kirchhoff's Current Law (KCL), which states that the total current entering a junction equals the total current leaving. Thus:
\[ I_T = I_1 + I_2 + I_3 \]
Substitute the known values:
\[ 12 \, \text{mA} = 3 \, \text{mA} + I_2 + 2 \, \text{mA} \]
Solve for \( I_2 \):
\[ I_2 = 12 \, \text{mA} - 3 \, \text{mA} - 2 \, \text{mA} = 7 \, \text{mA} \]
Therefore, the current through R2 is 7 mA.](https://content.bartleby.com/qna-images/question/f7186c19-f484-472d-921d-33d7fb24a522/66e60aec-4c56-4509-a8ec-737981aea4aa/eshoqs_processed.png)
Transcribed Image Text:**8. Determine the value of the current through R2 in the circuit shown.**
Given:
- \( I_T = 12 \, \text{mA} \)
- \( I_1 = 3 \, \text{mA} \)
- \( I_3 = 2 \, \text{mA} \)
**Diagram Explanation:**
The diagram illustrates a parallel circuit with three resistors: R1, R2, and R3. Current \( I_T \) enters the circuit and splits into three separate currents denoted as \( I_1 \), \( I_2 \), and \( I_3 \), flowing through R1, R2, and R3 respectively. The main current source \( I_T \) is given as 12 mA, with \( I_1 \) as 3 mA, and \( I_3 \) as 2 mA.
To determine the current through R2 (\( I_2 \)), use Kirchhoff's Current Law (KCL), which states that the total current entering a junction equals the total current leaving. Thus:
\[ I_T = I_1 + I_2 + I_3 \]
Substitute the known values:
\[ 12 \, \text{mA} = 3 \, \text{mA} + I_2 + 2 \, \text{mA} \]
Solve for \( I_2 \):
\[ I_2 = 12 \, \text{mA} - 3 \, \text{mA} - 2 \, \text{mA} = 7 \, \text{mA} \]
Therefore, the current through R2 is 7 mA.
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