### Educational Page Transcription: Electrical Measurements and Circuit Analysis#### Problem 2: Circuit Analysis with Voltmeter and AmmeterIn the circuit shown below, the voltmeter has an internal resistance of **100 kΩ** and the ammeter has a negligible resistance. (Show all needed calculations for parts a, b, c, d, & e below):**a. What is the equivalent resistance, \( R_p \), of the voltmeter & the resistor?****Calculations:**\[ \frac{1}{R_p} = \frac{1}{100\ k\Omega} + \frac{1}{200\ k\Omega} \]\[ R_p = \frac{100 \times 200}{100 + 200} = \frac{20000}{300} \approx 66.7\ k\Omega\]**b. How much current, \( I \), is the ammeter reading?****c. What is the voltmeter reading, \( V \)?****d. What is the \( \% E \) between the calculated value for \( R_p \) & the value given in the circuit diagram?****e. How can we reposition the voltmeter to reduce this error, & why will this position work?**##### Circuit Diagram:- **Components:** - Voltmeter (V) with internal resistance \( R_v = 100\ k\Omega \) - Ammeter (A) with negligible resistance - Resistor \( R = 200\ k\Omega \) - Power Supply: \( 50\ V \)- **Connection Description:** - The resistor \( R \) and the voltmeter are connected in parallel. - The ammeter is connected in series with the power supply.- **Visual Explanation:** - The power supply of 50 V is connected to the ammeter. - From the ammeter, the circuit branches into two parallel paths: one leading to the 200 kΩ resistor and the other to the 100 kΩ voltmeter. - These paths reconvene to complete the circuit back to the power supply.This diagram and the accompanying questions aim to enhance understanding of complex circuit behavior, particularly focusing on the interactions between voltmeter resistance, equivalent resistance, and current measurement.

a) Rp =R1R2/R1+R2I Rp = 100(200)/200+200 =200/3 =66.6 b) V= IRp I = V/Rp…

2. In the circuit shown below, the voltmeter has an internal resistance of 100 ks & the ammeter has a negligible resistance. (show all needed calculations for a, b, c, d, & e below). 100 ks a. What is the equivalent resistance, Rp, of the voltmeter & the V resistor? R A 〃 + = R₂ ने R₂ = 0.015 R₁ 100 200 200 ΚΩ R₂ = 0.015 = 66.7 ks 50 V b. How much current, I, is the ammeter reading? c. What is the voltmeter reading, V? d. What is the %E between the calculated value for R & the value given in the circuit diagram? e. How can we reposition the voltmeter to reduce this error, & why will this position work?

2. In the circuit shown below, the voltmeter has an internal resistance of 100 ks & the ammeter has a negligible resistance. (show all needed calculations for a, b, c, d, & e below). 100 ks a. What is the equivalent resistance, Rp, of the voltmeter & the V resistor? R A 〃 + = R₂ ने R₂ = 0.015 R₁ 100 200 200 ΚΩ R₂ = 0.015 = 66.7 ks 50 V b. How much current, I, is the ammeter reading? c. What is the voltmeter reading, V? d. What is the %E between the calculated value for R & the value given in the circuit diagram? e. How can we reposition the voltmeter to reduce this error, & why will this position work?

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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### Educational Page Transcription: Electrical Measurements and Circuit Analysis

#### Problem 2: Circuit Analysis with Voltmeter and Ammeter

In the circuit shown below, the voltmeter has an internal resistance of **100 kΩ** and the ammeter has a negligible resistance. (Show all needed calculations for parts a, b, c, d, & e below):

**a. What is the equivalent resistance, \( R_p \), of the voltmeter & the resistor?**

**Calculations:**
\[
\frac{1}{R_p} = \frac{1}{100\ k\Omega} + \frac{1}{200\ k\Omega}
\]
\[
R_p = \frac{100 \times 200}{100 + 200} = \frac{20000}{300} \approx 66.7\ k\Omega
\]

**b. How much current, \( I \), is the ammeter reading?**

**c. What is the voltmeter reading, \( V \)?**

**d. What is the \( \% E \) between the calculated value for \( R_p \) & the value given in the circuit diagram?**

**e. How can we reposition the voltmeter to reduce this error, & why will this position work?**

##### Circuit Diagram:
- **Components:**
- Voltmeter (V) with internal resistance \( R_v = 100\ k\Omega \)
- Ammeter (A) with negligible resistance
- Resistor \( R = 200\ k\Omega \)
- Power Supply: \( 50\ V \)
- **Connection Description:**
- The resistor \( R \) and the voltmeter are connected in parallel.
- The ammeter is connected in series with the power supply.
- **Visual Explanation:**
- The power supply of 50 V is connected to the ammeter.
- From the ammeter, the circuit branches into two parallel paths: one leading to the 200 kΩ resistor and the other to the 100 kΩ voltmeter.
- These paths reconvene to complete the circuit back to the power supply.

This diagram and the accompanying questions aim to enhance understanding of complex circuit behavior, particularly focusing on the interactions between voltmeter resistance, equivalent resistance, and current measurement.

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