Answered: What would be the current and voltage…

What would be the current and voltage passing through the 372 resistor when connected to a 12.0V battery? 490 370 750 R, 450

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Question

What would be the current and voltage passing through the
372 resistor when connected to a 12.0V battery?
490
sen
370
750
R,
450

Expert Solution

Step 1

The resistance R₄ and R₅ are connected in parallel and the equivalent resistance can be written as

$\begin{array}{l} \frac{1}{R_{45}} = \frac{1}{R_{4}} + \frac{1}{R_{5}} \\ \frac{1}{R_{45}} = \frac{1}{(45 Ω)} + \frac{1}{(37 Ω)} \\ \frac{1}{R_{45}} = \frac{82}{1665 Ω} \\ R_{45} = 20.3 Ω [1] \end{array}$

The resistance R₄₅ and R₆ are connected in series and the equivalent resistance can be written as

$\begin{array}{l} R_{456} = R_{45} + R_{6} \\ R_{456} = (20.3 Ω) + (75 Ω) \\ R_{456} = 95.3 Ω [2] \end{array}$

The resistance R₂ and R₃ are connected in series and the equivalent resistance can be written as

$\begin{array}{l} R_{23} = R_{2} + R_{3} \\ R_{23} = (49 Ω) + (51 Ω) \\ R_{23} = 100 Ω [3] \end{array}$

The resistance R₂₃ and R₄₅₆ are connected in parallel and the equivalent resistance can be written as

$\begin{array}{l} \frac{1}{R_{23456}} = \frac{1}{R_{23}} + \frac{1}{R_{456}} \\ \frac{1}{R_{23456}} = \frac{1}{(100 Ω)} + \frac{1}{(95.3 Ω)} \\ R_{23456} = 48.8 Ω [4] \end{array}$

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