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Why is static electricity considered to be a charge and not a current?
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Why is static electricity considered to be a charge and not a current?
Answer to Problem 1RQ
Static electricity refers to electrons that are immobile and not moving. Hence, it is a charge and not a current.
Explanation of Solution
Description:
The word static is used to express any quantity that is not moving or idle. Static electricity refers to electrons that are idle and immobile. Hence, it is considered to be a charge and not a current. A static charge can be either positive or negative depending upon the number of electrons. If an object has less electrons, it will have a positive charge; and if it has an excess of electrons, it will carry a negative charge.
Insulators are the only materials that hold the electrons motionless and keep them from flowing to a different position. Hence, electrostatic charges are built up on insulator materials. Electrostatic charges can be built up on a conductor only if it is electrically insulated from the surrounding objects.
Conclusion:
The reasons why static electricity is considered to be a charge and not a current has been explained.
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Chapter 3 Solutions
Delmar's Standard Textbook of Electricity (MindTap Course List)
- DUC 1. In Fig. 12-4, what are the functions of the VR1 and VR2? 2. In Fig. 12-4, what is the function of the VR3? VR₁ 50k C₁ R1 0.1 100k Carrier Input U₁ A741 PWM signal input R41k www Re 1k w C7 ± 10μT R7 100 ww =L H C4 2.2 H W82 Rs 51 3 10 U3 MC1496 C2 R2 U2 A741 22 0.1 100k VR2 50k VR3 100kr 14 C3 10μ 1k 0.1 4 5 6 12 m Re 10k R9 R102 3.9k 3.9k HHI C10 0.1 -0 +12V C11 R 0.02 100k +12 V Demodulated output C R11 R12 A741 0.33 10k 100k -12 V Ca 1μ C12 1500p PRODUC Fig. 12-4 PWM demodulator PRODUCTSarrow_forward10.37 Use mesh analysis to find currents I₁, I2, and I3 in the circuit of Fig. 10.82. ML 120-90° V 120 -30° V Figure 10.82 For Prob. 10.37. N N Z=80-135arrow_forward3. Find the phasor current I。 in the circuit shown below. Be aware of the direction markings. (15 pts) 1052 I 5057 ①520 Amps 2012 j5052arrow_forward
- 10.93 Figure 10.135 shows a Colpitts oscillator. Show that the ed oscillation frequency is 1 fo= 2π √√LCT where CTC₁C2/(C₁ + C₂). Assume R; >>> R₁ + Rf ww Vo L m C₂ C₁ 5 Xci Figure 10.135 A Colpitts oscillator; for Prob. 10.93. (Hint: Set the imaginary part of the impedance in the feedback circuit equal to zero.)arrow_forwardDetermine (a) the average and (b) rms values of the periodiccurrent waveform shown in Fig. P8.3.arrow_forward10.68 Find the Thevenin equivalent at terminals a-b in the circuit of Fig. 10.111. ML 6 sin 10t V 492 Figure 10.111 For Prob. 10.68. 5913 + 410 + -2 F 20 1H Vo obarrow_forward
- 10.79 For the op amp circuit in Fig. 10.122, obtain Vo. 5 cos 10³t V(+ Figure 10.122 For Prob. 10.79. 10 ΚΩ www 20 ΚΩ www 0.1 µF 40 ΚΩ 0.2 μFarrow_forward10.19 Obtain V, in Fig. 10.68 using nodal analysis. # ML ΖΩ j20 m 12/0° V 492 (+ ww www ' < ་ + V -j4 0.2V Figure 10.68 For Prob. 10.19.arrow_forward10.47 Determine i, in the circuit of Fig. 10.92, using the superposition principle. ML 10 sin(t -30°) V 1Ω www Figure 10.92 For Prob. 10.47. 96 F 202 www 24 V +) 2 H m io 2 cos 3t www 42arrow_forward
- 10.53 Use the concept of source transformation to find V, in the circuit of Fig. 10.97. 492 www -j30 j40 m + 20/0° V(+ j20 ΖΩ www -120 V ° Figure 10.97 For Prob. 10.53.arrow_forward2. Given you have a real valued signal with the following single sided baseband signal spectrum: ↑ ❘m(f)| A f=0 500 750 Sketch the frequency domain of |X(f)| given: a. x1(t) =m(t)cos(2**5000*) b. x2(t)=m(t)cos(2**600) Frequency (Hz)arrow_forwardwhat is deference between full Adder and Half?arrow_forward
- Delmar's Standard Textbook Of ElectricityElectrical EngineeringISBN:9781337900348Author:Stephen L. HermanPublisher:Cengage LearningElectricity for Refrigeration, Heating, and Air C...Mechanical EngineeringISBN:9781337399128Author:Russell E. SmithPublisher:Cengage Learning
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