for the first question I planned to calculate the current immediately after by I=emf/total R. (1)I’m not sure why In the solution it says that no current reaches R2 and that it short circuits so we only use R1. (2)Also in general if had a circuit with two branches and one branch had an open switch would current not flow at all in the whole circuit or only flow in the branch with no open switch?

Delmar's Standard Textbook Of Electricity
7th Edition
ISBN:9781337900348
Author:Stephen L. Herman
Publisher:Stephen L. Herman
Chapter17: Resistive-inductive Series Circuits
Section: Chapter Questions
Problem 2PA: You are a journeyman electrician working in an industrial plant. Your task is to connect an inductor...
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for the first question I planned to calculate the current immediately after by I=emf/total R. (1)I’m not sure why In the solution it says that no current reaches R2 and that it short circuits so we only use R1. (2)Also in general if had a circuit with two branches and one branch had an open switch would current not flow at all in the whole circuit or only flow in the branch with no open switch?
Example For a circuit shown in figure below & = 100 V, R₁5, R₂ = 102 and
C = 8μF.
w
R₂
I =
Figure 13: Example
. What is the initial battery current immediately after the switch S is closed?
• Solution: Immediately after the switch is closed, the capacitor is uncharged.
Thus, there is essentially no voltage drop across it, and it is like a wire. Thus,
resistor R₂ is short-circuited and so the battery current is
€
R
I =
R₁
E
R₁
100
5
= 20 A
• What is the battery current a long time after the switch S is closed?
• Solution: A long time after the current has been running through the circuit,
the capacitor will be fully charged and so no current can pass through it. Thus,
the current will flow through both R₁ and R₂ and will completely ignore the wire
with the capacitor on it. The current is then:
E
100
R₁ + R₂ 5 + 10
(22)
6.67 A
(23)
Transcribed Image Text:Example For a circuit shown in figure below & = 100 V, R₁5, R₂ = 102 and C = 8μF. w R₂ I = Figure 13: Example . What is the initial battery current immediately after the switch S is closed? • Solution: Immediately after the switch is closed, the capacitor is uncharged. Thus, there is essentially no voltage drop across it, and it is like a wire. Thus, resistor R₂ is short-circuited and so the battery current is € R I = R₁ E R₁ 100 5 = 20 A • What is the battery current a long time after the switch S is closed? • Solution: A long time after the current has been running through the circuit, the capacitor will be fully charged and so no current can pass through it. Thus, the current will flow through both R₁ and R₂ and will completely ignore the wire with the capacitor on it. The current is then: E 100 R₁ + R₂ 5 + 10 (22) 6.67 A (23)
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