Calculating Charge on Capacitor: Integration Help for a 5mA Constant Current

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Homework Help Overview

The discussion revolves around calculating the charge on a capacitor when a constant current of 5mA flows for a duration of 0.3 seconds. Participants are exploring the relationship between current and charge using calculus concepts.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to use integration to find the charge based on a constant current, questioning if their approach is correct. Other participants suggest that integration may not be necessary since the current is constant and propose a multiplication approach instead.

Discussion Status

Participants are actively engaging in clarifying the correct method to calculate charge under constant current conditions. Some guidance has been provided regarding the use of multiplication instead of integration, but no consensus has been reached on the final approach.

Contextual Notes

There is a mention of the original poster's uncertainty with calculus, which may affect their understanding of the problem. The discussion also highlights the assumption that the current is constant throughout the given time interval.

barry-
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hello all new to the forum i have a calculus problem wonder if anyone can help the charge (q) can be obtained by intergrating (i) with respect to time determine the the charge on capacitor as a result of current flowing for three hundredths of a second if (i) is a constant 5mA, my calculus is very rusty but have i got the equation right
5
∫ 5 di which i get the answer to 23.5 if I am doing it right.
0.300


Any help would be great thanks.
 
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barry- said:
hello all new to the forum i have a calculus problem wonder if anyone can help the charge (q) can be obtained by intergrating (i) with respect to time determine the the charge on capacitor as a result of current flowing for three hundredths of a second if (i) is a constant 5mA, my calculus is very rusty but have i got the equation right
5
∫ 5 di which i get the answer to 23.5 if I am doing it right.
0.300


Any help would be great thanks.

Welcome to the PF.

IF the current i is constant, you don't need to use integration. You only need to use integration if i varies with time...
 
ah ok so q = integral of (idt) is that right
 
barry- said:
ah ok so q = integral of (idt) is that right

Again, since i(t)=constant, you don't need the integral. You just multiply.

The fundamental equation for when the current is constant is

i(t) = ΔQ / Δt

So the change in charge with time is constant if I(t) is constant. What multiplication should you use to find the charge accumulated on the capacitor in a time Δt?
 
ok thanks for your help.
 

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