Explanation Given: The electromotive force is applied to an RC-series is 200 -volt. The resistance is 1000 ohms and the capacitance is 5 × 10 − 6 farad. Procedure used: 1. Write the circuit equation. 2. Find the integrating factor. 3. Obtain the final solution of differential equation. 4. Substitute the condition for obtain the constant value. Calculation: Consider the RC-series, R d q d t + 1 C q = E ( t ) (1) Here, d q d t is the current i , R is the resistance, C is the capacitance, q is the charge and E ( t ) is the electromotive force. Substitute 1000 for R , 5 × 10 − 6 for C and 200 for E ( t ) in equation (1). 1000 d q d t + 1 5 × 10 − 6 q = 200 d q d t + 200 q = 0.2 . (2) The solution of differential equation is, q ( I .F ) = ∫ 0.2 ( I .F ) d t + c (3) Obtain integrating factor of differential equation in equation (2). I .F= e ∫ 200 d t I .F= e 200 t Substitute e 200 t for I .F in equation (3) and integrate. q ( e 200 t ) = ∫ 0.2 ( e 200 t ) d t + c q ( e 200 t ) = 1 1000 e 200 t + c q = 1 1000 + c e − 200 t (4) Differentiate equation (4) with respect to t

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 2.7, Problem 32E

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The charge q(t) on the capacitor if i(0)=0 and the charge and current at t=0.05 sec and charge as t→∞.

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Chapter 2.7, Problem 31E

Chapter 2.7, Problem 33E

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The charge q ( t ) on the capacitor if i ( 0 ) = 0 and the charge and current at t = 0.05 sec and charge as t → ∞ .

Solution Summary: The author analyzes the electromotive force applied to an RC-series.

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The charge q(t) on the capacitor if i(0)=0 and the charge and current at t=0.05 sec and charge as t→∞.

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Chapter 2 Solutions