Explanation Calculation: The given circuit is shown in Figure 1. Figure 1 If the circuit has been connected for a long time, then it is considered to be in steady-state form. To determine the steady-state values, the capacitor is replaced by the open circuit and the inductor is replaced by the short circuit. The modified circuit diagram is shown in Figure 2. Figure 2 From the above diagram, the current across the current is zero because no current flows across the open circuit. The total current flow through the 2 k Ω resistor. The value of the current i R is calculated as, i R = 4 mA The conversion of k Ω into Ω is given by, 1 k Ω = 10 3 Ω The conversion of 2 k Ω into Ω is given by, 2 k Ω = 2 × 10 3 Ω The conversion of mA into A is given by, 1 mA = 10 − 3 A The conversion of 4 mA into mA is given by, 4 mA = 4 × 10 − 3 A The voltage across the resis

Electrical Engineering: Principles & Applications (7th Edition)

7th Edition

ISBN: 9780134484143

Author: Allan R. Hambley

Publisher: PEARSON

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Chapter 4, Problem 4.27P

To determine

The value of the vC and iR.

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Chapter 4, Problem 4.26P

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Chapter 4 Solutions

Electrical Engineering: Principles & Applications (7th Edition)

Ch. 4 - Suppose we have a capacitance C discharging...Ch. 4 - The dielectric materials used in real capacitors...Ch. 4 - The initial voltage across the capacitor shown in...Ch. 4 - A 100F capacitance is initially charged to 1000 V....Ch. 4 - At t = 0, a charged 10{ F capacitance is connected...Ch. 4 - At time t1 , a capacitance C is charged to a...Ch. 4 - Given an initially charged capacitance that begins...Ch. 4 - The initial voltage across the capacitor shown in...Ch. 4 - In physics, the half-life is often used to...Ch. 4 - We know that a 50F capacitance is charged to an...

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The value of the v C and i R .

Solution Summary: The author calculates the value of the given circuit in Figure 1. The capacitor is replaced by an open circuit and the inductor by a short circuit.

The value of the vC and iR.

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Chapter 4 Solutions