### Understanding the Effect of Switch Operation on Current in a CircuitWhen analyzing the behavior of electrical circuits, it is important to understand how different circuit configurations affect the current flow. Consider the given circuit diagram:![Circuit Diagram](link_to_image)This circuit consists of a power source, an ammeter (A), a resistor of 15 ohms, and a parallel combination of a 20-ohm resistor and a 60-ohm resistor. A switch, S, is used to control whether the 60-ohm resistor is included in the circuit.#### Scenario Analysis:1. **Switch S is Open:** - When the switch S is open, the 60-ohm resistor is not part of the circuit. - The total resistance in the circuit is just the sum of the 15-ohm resistor and the 20-ohm resistor, R_total_open = 15Ω + 20Ω = 35Ω. - Given that the current through the ammeter (A) is 2.0 A when S is open, we can use Ohm's law (V = IR) to find the voltage across the circuit: V = 2.0 A * 35Ω = 70 V.2. **Switch S is Closed:** - When the switch S is closed, the 60-ohm resistor comes into play, creating a parallel network with the 20-ohm resistor. - The equivalent resistance of the parallel network (R_parallel) is given by the formula for resistors in parallel: \( R_{\text{parallel}} = \left( \frac{1}{20Ω} + \frac{1}{60Ω} \right)^{-1} \) \[ R_{\text{parallel}} = \left( \frac{1}{20} + \frac{1}{60} \right)^{-1} = \left( \frac{3}{60} + \frac{1}{60} \right)^{-1} = \left( \frac{4}{60} \right)^{-1} = \frac{60}{4} = 15Ω \] - The total resistance in the circuit now is the series combination of the 15Ω resistor and the equivalent parallel resistance: R_total_closed = 15Ω + 15Ω = 30Ω. - Using Ohm's law again with the same voltage source, we find

Given data: R1=15 ohm R2=20 ohm R3=60 ohm Current I=2A

When the switch S is open in the circuit shown, the reading on the ammeter A is 2.0 A. When the switch is closed, the reading on the ammeter is A ww 20 nE 602 doubled O increased slightly but not doubled O decreased slightly but not halved O halved

When the switch S is open in the circuit shown, the reading on the ammeter A is 2.0 A. When the switch is closed, the reading on the ammeter is A ww 20 nE 602 doubled O increased slightly but not doubled O decreased slightly but not halved O halved

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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Question

### Understanding the Effect of Switch Operation on Current in a Circuit

When analyzing the behavior of electrical circuits, it is important to understand how different circuit configurations affect the current flow. Consider the given circuit diagram:

![Circuit Diagram](link_to_image)

This circuit consists of a power source, an ammeter (A), a resistor of 15 ohms, and a parallel combination of a 20-ohm resistor and a 60-ohm resistor. A switch, S, is used to control whether the 60-ohm resistor is included in the circuit.

#### Scenario Analysis:

1. **Switch S is Open:**
- When the switch S is open, the 60-ohm resistor is not part of the circuit.
- The total resistance in the circuit is just the sum of the 15-ohm resistor and the 20-ohm resistor, R_total_open = 15Ω + 20Ω = 35Ω.
- Given that the current through the ammeter (A) is 2.0 A when S is open, we can use Ohm's law (V = IR) to find the voltage across the circuit: V = 2.0 A * 35Ω = 70 V.

2. **Switch S is Closed:**
- When the switch S is closed, the 60-ohm resistor comes into play, creating a parallel network with the 20-ohm resistor.
- The equivalent resistance of the parallel network (R_parallel) is given by the formula for resistors in parallel: \( R_{\text{parallel}} = \left( \frac{1}{20Ω} + \frac{1}{60Ω} \right)^{-1} \)
\[
R_{\text{parallel}} = \left( \frac{1}{20} + \frac{1}{60} \right)^{-1} = \left( \frac{3}{60} + \frac{1}{60} \right)^{-1} = \left( \frac{4}{60} \right)^{-1} = \frac{60}{4} = 15Ω
\]
- The total resistance in the circuit now is the series combination of the 15Ω resistor and the equivalent parallel resistance: R_total_closed = 15Ω + 15Ω = 30Ω.
- Using Ohm's law again with the same voltage source, we find

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