[Q13] The figure shows R₁ = 200 , R₂ = 100 , and the ideal batteries (no internal resistance) have emfs of &₁ = 7.0 V, E₂ = 6.0 V, and E3 = 5.0 V. Find the current through the resistor R₂ by using Kirchoff's laws on loop A. V = emf (1-e²+/RC) V = erf(1-2.718-1/100n()) 11.ov = (1-0,99005086) 11.or=(0.00994914²) 11.0v) 0.00994914) a E2 R₁ E ܐ E₁ #190² R₂ Va-V₂=E₁ + ₂ + Ez X -4 = √₂ + ²3 =-10.99 [C=10.99€ C
[Q13] The figure shows R₁ = 200 , R₂ = 100 , and the ideal batteries (no internal resistance) have emfs of &₁ = 7.0 V, E₂ = 6.0 V, and E3 = 5.0 V. Find the current through the resistor R₂ by using Kirchoff's laws on loop A. V = emf (1-e²+/RC) V = erf(1-2.718-1/100n()) 11.ov = (1-0,99005086) 11.or=(0.00994914²) 11.0v) 0.00994914) a E2 R₁ E ܐ E₁ #190² R₂ Va-V₂=E₁ + ₂ + Ez X -4 = √₂ + ²3 =-10.99 [C=10.99€ C
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 13**
The figure shows \( R_1 = 200 \, \Omega \), \( R_2 = 100 \, \Omega \), and the ideal batteries (no internal resistance) have emfs of \( \varepsilon_1 = 7.0 \, \text{V} \), \( \varepsilon_2 = 6.0 \, \text{V} \), and \( \varepsilon_3 = 5.0 \, \text{V} \). Find the current through the resistor \( R_2 \) by using Kirchhoff's laws on loop A.
**Circuit Diagram:**
- The diagram is a simple loop circuit.
- It consists of three batteries with emfs \( \varepsilon_1 \), \( \varepsilon_2 \), \( \varepsilon_3 \).
- Two resistors \( R_1 \) and \( R_2 \) are connected in the loop.
- Points A and B are marked in the circuit.
**Equations:**
- Using Kirchhoff's law:
\[
V_{ab} - V_b = \varepsilon_1 + \varepsilon_2 + \varepsilon_3
\]
- Rearrange:
\[
V_{ab} - V_b - \varepsilon_1 = \varepsilon_2 + \varepsilon_3
\]
- Substitute values:
\[
V_{ab} = 6.0 \, \text{V} + 5.0 \, \text{V}
\]
\[
V_{ab} = 11.0 \, \text{V}
\]
**Calculations:**
- Voltage equation:
\[
V = \text{emf} (1 - e^{-t/100RC})
\]
- Substitute known values:
\[
11.0 \, \text{V} = (1 - 0.99005086)
\]
\[
11.0 \, \text{V} = (0.009994914)
\]
- Current equation:
\[
\frac{i}{c} = \left( 11.0 \, \text{V} \right) (0.009994914)
\]
\[
i = C =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F18d0a7dc-afe1-42e7-9349-84d2cfc69c53%2F8590ff55-e446-4b2c-9ea7-ad11f1e8acc2%2Fltln1q9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 13**
The figure shows \( R_1 = 200 \, \Omega \), \( R_2 = 100 \, \Omega \), and the ideal batteries (no internal resistance) have emfs of \( \varepsilon_1 = 7.0 \, \text{V} \), \( \varepsilon_2 = 6.0 \, \text{V} \), and \( \varepsilon_3 = 5.0 \, \text{V} \). Find the current through the resistor \( R_2 \) by using Kirchhoff's laws on loop A.
**Circuit Diagram:**
- The diagram is a simple loop circuit.
- It consists of three batteries with emfs \( \varepsilon_1 \), \( \varepsilon_2 \), \( \varepsilon_3 \).
- Two resistors \( R_1 \) and \( R_2 \) are connected in the loop.
- Points A and B are marked in the circuit.
**Equations:**
- Using Kirchhoff's law:
\[
V_{ab} - V_b = \varepsilon_1 + \varepsilon_2 + \varepsilon_3
\]
- Rearrange:
\[
V_{ab} - V_b - \varepsilon_1 = \varepsilon_2 + \varepsilon_3
\]
- Substitute values:
\[
V_{ab} = 6.0 \, \text{V} + 5.0 \, \text{V}
\]
\[
V_{ab} = 11.0 \, \text{V}
\]
**Calculations:**
- Voltage equation:
\[
V = \text{emf} (1 - e^{-t/100RC})
\]
- Substitute known values:
\[
11.0 \, \text{V} = (1 - 0.99005086)
\]
\[
11.0 \, \text{V} = (0.009994914)
\]
- Current equation:
\[
\frac{i}{c} = \left( 11.0 \, \text{V} \right) (0.009994914)
\]
\[
i = C =
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