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A student in PY106 decides to build a rather strange circuit using two batteries (9 V and 3 V) and two resistors (10 and 2 Q) as shown in the figure below. They want to calculate the current flowing around this circuit (in the direction shown in the figure) but are not quite sure how to start. Don't worry, Kirchhoff's Rules to the rescue! Assuming we travel around the circuit in the clockwise direction (starting at point a and moving up through the 9 V battery), how would we write Kirchhoff's Loop Rule? 9 V 3 V I d.

A student in PY106 decides to build a rather strange circuit using two batteries (9 V and 3 V) and two resistors (10 and 2 Q) as shown in the figure below. They want to calculate the current flowing around this circuit (in the direction shown in the figure) but are not quite sure how to start. Don't worry, Kirchhoff's Rules to the rescue! Assuming we travel around the circuit in the clockwise direction (starting at point a and moving up through the 9 V battery), how would we write Kirchhoff's Loop Rule? 9 V 3 V I d.

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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**Circuit Analysis Using Kirchhoff's Loop Rule** A student in PY106 decides to build a rather strange circuit using two batteries (9 V and 3 V) and two resistors (1 Ω and 2 Ω) as shown in the figure below. They want to calculate the current flowing around this circuit (in the direction shown in the figure) but are not quite sure how to start. Don't worry, Kirchhoff's Rules to the rescue! **Figure Description:** - The circuit is comprised of four points labeled a, b, c, and d. - Starting at point a, the circuit moves upwards through a 9 V battery. - The circuit then moves horizontally from point b to point c through a 1 Ω resistor, marked with current I flowing to the right. - Continuing from point c, the circuit moves horizontally through a 2 Ω resistor with current I also flowing to the right. - From point c, the circuit moves downward through a 3 V battery. - Finally, the circuit completes the loop from point d back to point a. **Assumption:** Assuming we travel around the circuit in the clockwise direction (starting at point a and moving up through the 9 V battery), how would we write Kirchhoff's Loop Rule? To solve the circuit using Kirchhoff’s Loop Rule, we sum the voltage gains and drops around the loop: - Gain of 9 V from the 9 V battery - Drop across the 1 Ω resistor: 1Ω * I - Drop across the 2 Ω resistor: 2Ω * I - Drop of 3 V from the 3 V battery Applying Kirchhoff's Loop Rule: \[ 9 \, \text{V} - 1 \, \Omega \cdot I - 2 \, \Omega \cdot I - 3 \, \text{V} = 0 \] This equation can be solved for the current I flowing in the circuit.
Transcribed Image Text:**Circuit Analysis Using Kirchhoff's Loop Rule** A student in PY106 decides to build a rather strange circuit using two batteries (9 V and 3 V) and two resistors (1 Ω and 2 Ω) as shown in the figure below. They want to calculate the current flowing around this circuit (in the direction shown in the figure) but are not quite sure how to start. Don't worry, Kirchhoff's Rules to the rescue! **Figure Description:** - The circuit is comprised of four points labeled a, b, c, and d. - Starting at point a, the circuit moves upwards through a 9 V battery. - The circuit then moves horizontally from point b to point c through a 1 Ω resistor, marked with current I flowing to the right. - Continuing from point c, the circuit moves horizontally through a 2 Ω resistor with current I also flowing to the right. - From point c, the circuit moves downward through a 3 V battery. - Finally, the circuit completes the loop from point d back to point a. **Assumption:** Assuming we travel around the circuit in the clockwise direction (starting at point a and moving up through the 9 V battery), how would we write Kirchhoff's Loop Rule? To solve the circuit using Kirchhoff’s Loop Rule, we sum the voltage gains and drops around the loop: - Gain of 9 V from the 9 V battery - Drop across the 1 Ω resistor: 1Ω * I - Drop across the 2 Ω resistor: 2Ω * I - Drop of 3 V from the 3 V battery Applying Kirchhoff's Loop Rule: \[ 9 \, \text{V} - 1 \, \Omega \cdot I - 2 \, \Omega \cdot I - 3 \, \text{V} = 0 \] This equation can be solved for the current I flowing in the circuit.
**Question:** Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. **Options:** a) \(9V - I(1\Omega) - I(2\Omega) + 3V = 0\) b) \(-9V - I(1\Omega) - I(2\Omega) + 3V = 0\) c) \(9V - I(1\Omega) - I(2\Omega) - 3V = 0\) d) \(9V + I(1\Omega) + I(2\Omega) - 3V = 0\) e) \(9V + I(1\Omega) + I(2\Omega) + 3V = 0\) f) \(-9V + I(1\Omega) + I(2\Omega) + 3V = 0\) g) None of the above **Diagram:** The image also includes a segment of an electric circuit with points labeled "a" and "d" and a current \(I\) flowing through a branch. This setup illustrates the direction of current flow and the application of voltages in the circuit analysis problem.
Transcribed Image Text:**Question:** Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. **Options:** a) \(9V - I(1\Omega) - I(2\Omega) + 3V = 0\) b) \(-9V - I(1\Omega) - I(2\Omega) + 3V = 0\) c) \(9V - I(1\Omega) - I(2\Omega) - 3V = 0\) d) \(9V + I(1\Omega) + I(2\Omega) - 3V = 0\) e) \(9V + I(1\Omega) + I(2\Omega) + 3V = 0\) f) \(-9V + I(1\Omega) + I(2\Omega) + 3V = 0\) g) None of the above **Diagram:** The image also includes a segment of an electric circuit with points labeled "a" and "d" and a current \(I\) flowing through a branch. This setup illustrates the direction of current flow and the application of voltages in the circuit analysis problem.
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