Circuit City

Electricity and Magnetism Level 2

Circuits are powered by a 9 volt battery and current flows through a combination of resisters as shown in diagram. What is the current flowing in branch line F-C in amperes? The resisters are labeled in units of ohms. Provide your answer to the nearest 1/1000th.


The answer is 0.177.

A Former Brilliant Member by A Former Brilliant Member

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2 solutions

K T
Jan 31, 2021

Parallel resistors combine as R = R 1 R 2 R 1 + R 2 R=\frac{R_1R_2}{R_1+R_2} .

Therefore R A B = 5 × 8 5 + 8 = 40 13 R_{AB}= \frac{5×8}{5+8}=\frac{40}{13} and R D E = 6 × 9 6 + 9 = 18 5 R_{DE}= \frac{6×9}{6+9}=\frac{18}{5} . Between F and C we have 5 + 18 5 = 43 5 5+\frac{18}{5}=\frac{43}{5} in the lower branch and 15 in the upper branch (R4). Combined this is 645 118 \frac{645}{118}

The total resistance R = 10 + 40 13 + 645 118 = 28445 1534 R=10+\frac{40}{13}+\frac{645}{118}=\frac{28445}{1534} and so the total current equals I = V / R = 13806 28445 I=V/R=\frac{13806}{28445}

The current splits according to the inverse of the resistance, so through R4 flows 43 118 I \frac{43}{118} I and through R5 flows 75 118 I \frac{75}{118} I .

The requested answer therefore is 43 118 × 13806 28445 = 296829 1678255 ≈ 0.17687 A \frac{43}{118}× \frac{13806}{28445}=\frac{296829}{1678255}\approx 0.17687 A

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Should be listed as an E&M prob instead of Classical Mechanics.

tom engelsman - 4 months, 1 week ago

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Thanks for spotting that. Will change.

A Former Brilliant Member - 4 months, 1 week ago

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