Question - Determining Currents in a Circuit with a Voltage Source and Multiple Resistors
Solution:
Given \( V = 6V \), \( R1 = 1k\Omega \), \( R2 = 6k\Omega \), \( R3 = 6k\Omega \), \( R4 = 1k\Omega \), \( R5 = 10k\Omega \), \( R6 = 8k\Omega \), \( R7 = 2k\Omega \).
First, find the equivalent resistance for resistors in series and parallel.
Since \( R4 \) and \( R5 \) are in series, their equivalent resistance, \( R_{45} \), is:
\( R_{45} = R4 + R5 = 1k\Omega + 10k\Omega = 11k\Omega \)
Resistors \( R3 \), \( R_{45} \), and \( R6 \) are parallel, which means their combined equivalent resistance, \( R_{3456} \), is found using the formula:
\[
\frac{1}{R_{3456}} = \frac{1}{R3} + \frac{1}{R_{45}} + \frac{1}{R6} = \frac{1}{6k\Omega} + \frac{1}{11k\Omega} + \frac{1}{8k\Omega}
\]
Solving for \( R_{3456} \):
\[
R_{3456} = \frac{1}{(\frac{1}{6} + \frac{1}{11} + \frac{1}{8}) k\Omega^{-1}} \approx 2.97k\Omega
\]
Now, \( R2 \), \( R_{3456} \), and \( R7 \) are in series, so the total resistance \( R_{T} \) is:
\( R_{T} = R1 + R2 + R_{3456} + R7 = 1k\Omega + 6k\Omega + 2.97k\Omega + 2k\Omega = 11.97k\Omega \)
Using Ohm's law \( V = IR \), find the total current \( I \):
\( I = \frac{V}{R_{T}} = \frac{6V}{11.97k\Omega} \approx 0.501mA \)
Current \( I \) flows through \( R1 \), \( R2 \), \( R_{3456} \), and \( R7 \), so \( I_{R1} = I_{R2} = I_{R7} = 0.501mA \).
The voltage across \( R_{3456} \), \( V_{3456} \), is:
\( V_{3456} = I_{R_{3456}} \times R_{3456} = 0.501mA \times 2.97k\Omega \approx 1.49V \)
This voltage is the same across \( R3 \), \( R4 \), and \( R5 \) in parallel. Use Ohm's law to find the current through \( R3 \), \( I_{R3} \):
\( I_{R3} = \frac{V_{3456}}{R3} = \frac{1.49V}{6k\Omega} \approx 0.248mA \approx 248\mu A \)
Therefore, the currents through \( R1 \) (which is equivalent to \( I \)) and \( R3 \) are approximately \( 501\mu A \) and \( 248\mu A \) respectively.
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