Explanation Given data: The internal resistance of voltmeter is 100 k Ω . Refer to Figure given in the textbook. Calculation: In the given circuit, the voltage source 36 V with series resistance 12 k Ω is converted into current source 3 mA with the parallel resistance 12 k Ω using source transformation method as shown in Figure 1. That is, the current source is calculated by using Ohm’s law as follows. current source = 36 V 12 k Ω = 36 V 12 × 10 3 Ω { ∵ 1 k = 10 3 } = 3 × 10 − 3 A { ∵ 1 A = 1 V 1 Ω } = 3 mA { ∵ 1 m = 10 − 3 } Figure 1 shows the modified circuit diagram. In Figure 1, the two current source are in opposite direction. Therefore, the resultant current source is calculated as follows. current source = 18 mA − 3 mA = 15 mA Since the current source is positive, then the current direction of current source directed upwards as shown in Figure 2. In Figure 1, the resistor 12 k Ω and 60 k Ω are connected in parallel. Therefore, the equivalent resistance for parallel connection is calculated as follows. 12 k Ω | | 60 k Ω = 12 k Ω × 60 k Ω 12 k Ω + 60 k Ω = 10 k Ω Figure 2 shows the reduced circuit diagram

10th Edition

ISBN: 9780133760033

Author: James W. Nilsson, Susan Riedel

Publisher: PEARSON

See similar textbooks

Concept explainers

Electrical Heating Unit

The conversion of electrical energy to heat energy is known as electric heating. The heating system that converts the electrical energy to heat energy for various application is called electrical heating unit. Electricity is employed as the firing mechan…

Videos

Question

Chapter 4.10, Problem 18AP

To determine

Find the voltmeter reading across the terminals A-B of the circuit.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 4.10, Problem 17AP

Chapter 4.11, Problem 19AP

Students have asked these similar questions

Choose the appropriate answer 1) Maximum dimension of antenna is 0.5m and operating frequency is 9 GHz, thus the radius of reactive near field region is 0.562m 1.265m 2.526m 3.265m 2) If distance between transmitter and receiver is 2km and the signal carrier frequency is 300kHz Rapidly time-varying fields DC field Quasi-static field None 3) The polarization mismatch factor for horizontal polarization wave incident on +z axis is is if the antenna polarization is circular 0.5 зав 0.707 1 4) Ez 0 and Hz #0 (HE modes): This is the case when neither E nor H field is transverse to the direction of wave propagation. They are sometimes referred to as TEM hybrid modes TM TE 5) The normalized radiation intensity of an antenna is represented by: U(6)=cos²(0) cos2 (30), w/s Half-power beamwidth HPBW is...... 28.75 10 0 14.3

Choose the best answer of the following: 1- quasi-static electromagnetic field is the a) low frequency b)high frequency c) time independent d) none of the above 2- Displacement current is taken to be negligible (compared to the conduction current) if a) σ>>wε b)σ << wɛ c) σ =0 d) (a and c) 3- The transmission line act as inductor when it terminated by: a) Open circuit load b) short circuit load c)matched load d)none of the above 4- The scattering aperture equals to the effective aperture when the antenna is: a) Complex conjugate matching b) short circuit c) open circuit d) none of the above 5- The isotropic point source has directivity of: a) Infinity b)1 c) 0 d)1.5

I selected a DC-DC converter capable of delivering 120 VDC from a 600 VDC input. When I reached out to the manufacturer, they asked for the total power consumption the converter would need to handle.To estimate this, I calculated the power requirements for the components that will use the 120 VDC supply: interior lighting, end lights, and buzzers. The breakdown is as follows:- Light Bulbs: 16 bulbs at 10 W each = 160 W- Buzzers: 2 buzzers at 5 W each = 10 W- End Lights: 2 lights at 15 W each = 30 W This results in a total estimated power demand of 200 W.My concern is whether I should request a higher wattage rating for the converter to provide sufficient tolerance and ensure the system operates efficiently without risking an overload. Note: The DC power system is designed specifically for a trolley

Chapter 4 Solutions

Electric Circuits (10th Edition)

Ch. 4.2 - a) For the circuit shown, use the node-voltage...Ch. 4.2 - Use the node-voltage method to find v in the...Ch. 4.3 - Use the node-voltage method to find the power...Ch. 4.4 - Use the node-voltage method to find vo in the...Ch. 4.4 - Use the node-voltage method to find v in the...Ch. 4.4 - Use the node-voltage method to find v1 in the...Ch. 4.5 - Use the mesh-current method to find (a) the power...Ch. 4.6 - Determine the number of mesh-current equations...Ch. 4.6 - Use the mesh-current method to find vo in the...Ch. 4.7 - Use the mesh-current method to find the power...

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, electrical-engineering and related others by exploring similar questions and additional content below.