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t-0 Figure 1: RL Circuit 1. In an RL series circuit containing only a resistor and an inductor, Kirchhop's second law states that the sum of the voltage drop across the inductor(L di/dt) and the voltage drop across the resistor (iR) is the same as the impressed voltage (E(t)) on the circuit. The current i(t), called the response of the system, is governed by a differential equation di L dt + Ri = E(t) where L and R are constants known as inductance and the resistance respectively. Based on the above equation, solve the following: (a) A 12-volt battery is connected to a series in which the inductance is 0.5 henry and the resistance is 10 ohms. Determine the current i if the initial current is zero. Then determine the steady-state current for the system.

t-0 Figure 1: RL Circuit 1. In an RL series circuit containing only a resistor and an inductor, Kirchhop's second law states that the sum of the voltage drop across the inductor(L di/dt) and the voltage drop across the resistor (iR) is the same as the impressed voltage (E(t)) on the circuit. The current i(t), called the response of the system, is governed by a differential equation di L dt + Ri = E(t) where L and R are constants known as inductance and the resistance respectively. Based on the above equation, solve the following: (a) A 12-volt battery is connected to a series in which the inductance is 0.5 henry and the resistance is 10 ohms. Determine the current i if the initial current is zero. Then determine the steady-state current for the system.

Delmar's Standard Textbook Of Electricity
Delmar's Standard Textbook Of Electricity
7th Edition
ISBN:9781337900348
Author:Stephen L. Herman
Publisher:Stephen L. Herman
Chapter23: Resistive-inductive-capacitive Series Circuits
Section: Chapter Questions
Problem 6RQ
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t-0 R Figure 1: RL Circuit In an RL series circuit containing only a resistor and an inductor, Kirchhop's second law states that the sum of the voltage drop across the inductor(L di/dt) and the voltage drop across the resistor (iR) is the same as the impressed voltage (E(t)) on the circuit. The current i(t), called the response of the system, is governed by a differential equation 1. di L+ Ri = E(t) dt where L and R are constants known as inductance and the resistance respectively. Based on the above equation, solve the following: (a) A 12-volt battery is connected to a series in which the inductance is 0.5 henry and the resistance is 10 ohms. Determine the current i if the initial current is A zero. Then determine the steady-state current for the system. Ans: i(t) = (1 – e-20t) W-
Transcribed Image Text:t-0 R Figure 1: RL Circuit In an RL series circuit containing only a resistor and an inductor, Kirchhop's second law states that the sum of the voltage drop across the inductor(L di/dt) and the voltage drop across the resistor (iR) is the same as the impressed voltage (E(t)) on the circuit. The current i(t), called the response of the system, is governed by a differential equation 1. di L+ Ri = E(t) dt where L and R are constants known as inductance and the resistance respectively. Based on the above equation, solve the following: (a) A 12-volt battery is connected to a series in which the inductance is 0.5 henry and the resistance is 10 ohms. Determine the current i if the initial current is A zero. Then determine the steady-state current for the system. Ans: i(t) = (1 – e-20t) W-
The voltage drop across a capacitor with capacitance C is given by q(t)/C, where q is the charge on the capacitor. Hence, for the RC series circuit, Kirchhop's second law gives 2. 1 Ri + 79 E(t). But current i and charge q are related by i = dq/dt. So, the above equation becomes dą R- 1 E(t). dt T9 Use this differential equation to solve the following: (b) Let R = 10 ohms and C = 0.1 farad. Let E(t) be exponentially decaying, say, E(t) = 30e * volts. Assuming q(0) = 0, find and graph q(t). At what time does q(t) reach a maximum? What is that maximum charge? Ans: q(t) = 1.5(et -e-3t); q'(t) = 0 gives tm = 0.549; qm = 0.577 coulomb
Transcribed Image Text:The voltage drop across a capacitor with capacitance C is given by q(t)/C, where q is the charge on the capacitor. Hence, for the RC series circuit, Kirchhop's second law gives 2. 1 Ri + 79 E(t). But current i and charge q are related by i = dq/dt. So, the above equation becomes dą R- 1 E(t). dt T9 Use this differential equation to solve the following: (b) Let R = 10 ohms and C = 0.1 farad. Let E(t) be exponentially decaying, say, E(t) = 30e * volts. Assuming q(0) = 0, find and graph q(t). At what time does q(t) reach a maximum? What is that maximum charge? Ans: q(t) = 1.5(et -e-3t); q'(t) = 0 gives tm = 0.549; qm = 0.577 coulomb
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