Circuit Theory
... which is also Vout. KCL says the current leaving resistor R1 must equal the current entering R2, or IR1 = IR2, so we can write Vout = IR1 R2. KVL says the voltage around the loop including the battery and both resistors is 0, therefore Vin = VR1 + Vout, or Vin = IR1 R1 + IR1 R2. Thus, IR1 = Vin / (R ...
... which is also Vout. KCL says the current leaving resistor R1 must equal the current entering R2, or IR1 = IR2, so we can write Vout = IR1 R2. KVL says the voltage around the loop including the battery and both resistors is 0, therefore Vin = VR1 + Vout, or Vin = IR1 R1 + IR1 R2. Thus, IR1 = Vin / (R ...
1000BASE-T and 10/100/1000BASE-T Copper
... Mod-Def 1 is the clock line of two wire serial interface for serial ID Mod-Def 2 is the data line of two wire serial interface for serial ID 4) RX_LOS (Loss of Signal): LVTTL compatible with a maximum voltage of Host_Vcc. RX_LOS can enabled or disabled (Refer to Ordering information),RX_LOS is not u ...
... Mod-Def 1 is the clock line of two wire serial interface for serial ID Mod-Def 2 is the data line of two wire serial interface for serial ID 4) RX_LOS (Loss of Signal): LVTTL compatible with a maximum voltage of Host_Vcc. RX_LOS can enabled or disabled (Refer to Ordering information),RX_LOS is not u ...
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... 1. Label the N node voltages. The node voltages are defined positive with respect to a common point (i.e., the reference node) in the circuit generally designated as the ground (V = 0). 2. Apply KCL at each node in terms of node voltages. a. Use KCL to write a current balance at N-1 of the N nodes ...
... 1. Label the N node voltages. The node voltages are defined positive with respect to a common point (i.e., the reference node) in the circuit generally designated as the ground (V = 0). 2. Apply KCL at each node in terms of node voltages. a. Use KCL to write a current balance at N-1 of the N nodes ...
Load Characteristics
... power factor and reactive power used etc. • They can also support time-of-day billing, for example, recording the amount of energy used during onpeak and off-peak hours. ...
... power factor and reactive power used etc. • They can also support time-of-day billing, for example, recording the amount of energy used during onpeak and off-peak hours. ...
EE369 POWER SYSTEM ANALYSIS
... Balanced 3 Phase () Systems A balanced 3 phase () system has: – three voltage sources with equal magnitude, but with an angle shift of 120, – equal loads on each phase, – equal impedance on the lines connecting the generators to the loads. ...
... Balanced 3 Phase () Systems A balanced 3 phase () system has: – three voltage sources with equal magnitude, but with an angle shift of 120, – equal loads on each phase, – equal impedance on the lines connecting the generators to the loads. ...
X/R Ratio - Powell Industries
... of any load on the system. Since generators, transformers and transmission lines are generally quite highly inductive, the X/R ratio is generally significantly above unity in any utility or industrial power system. Why is the X/R ratio important? Its importance is that it affects the level of short ...
... of any load on the system. Since generators, transformers and transmission lines are generally quite highly inductive, the X/R ratio is generally significantly above unity in any utility or industrial power system. Why is the X/R ratio important? Its importance is that it affects the level of short ...
CHAPTER 7 : EFFECT OF TEMPERATURE UPON RESISTANCE
... A series R-L circuit of resistance of 25 Ω and inductance of 0.1 H, is connected to a 250-V, 50-Hz, supply. Calculate the (a) inductive reactance, (b) impedance, (c) current, (d) voltage across the resistive component, (e) voltage across the inductive component, (f) phase angle. [Answer: (a) 31.42 Ω ...
... A series R-L circuit of resistance of 25 Ω and inductance of 0.1 H, is connected to a 250-V, 50-Hz, supply. Calculate the (a) inductive reactance, (b) impedance, (c) current, (d) voltage across the resistive component, (e) voltage across the inductive component, (f) phase angle. [Answer: (a) 31.42 Ω ...
Solving the above equation we get
... The phase angle of the mid point voltage is half the load angle always. Also the mid point voltage and current are in phase, i.e., the power factor at this point is unity. The variation in the magnitude of voltage with changes in load angle is maximum at the mid point. The voltage at this point decr ...
... The phase angle of the mid point voltage is half the load angle always. Also the mid point voltage and current are in phase, i.e., the power factor at this point is unity. The variation in the magnitude of voltage with changes in load angle is maximum at the mid point. The voltage at this point decr ...
How Does Return Current Really Return?
... looks like some positive charge is added to the top conduc50Ω coax cable, 10’ in length. The onetor and some negative charge pushed out from the bottom way time delay is about 15 nsec. If we conductor. From the outside, it looks like current has gone launch a 1 V signal into the line, the curthrough ...
... looks like some positive charge is added to the top conduc50Ω coax cable, 10’ in length. The onetor and some negative charge pushed out from the bottom way time delay is about 15 nsec. If we conductor. From the outside, it looks like current has gone launch a 1 V signal into the line, the curthrough ...
GATE 2008 Electrical Engineering
... is the input to a Linear Time invariant system whose impulse response h(t) = sinc ( ) where is a real constant. If min (a, ) denotes the minimum of ∝ and , and similarly max (∝, ) denotes the maximum of ∝ and , and K is a constant, which one of the following statements is true about the output of th ...
... is the input to a Linear Time invariant system whose impulse response h(t) = sinc ( ) where is a real constant. If min (a, ) denotes the minimum of ∝ and , and similarly max (∝, ) denotes the maximum of ∝ and , and K is a constant, which one of the following statements is true about the output of th ...
Standing wave ratio
In radio engineering and telecommunications, standing wave ratio (SWR) is a measure of impedance matching of loads to the characteristic impedance of a transmission line or waveguide. Impedance mismatches result in standing waves along the transmission line, and SWR is defined as the ratio of the partial standing wave's amplitude at an antinode (maximum) to the amplitude at a node (minimum) along the line.The SWR is usually thought of in terms of the maximum and minimum AC voltages along the transmission line, thus called the voltage standing wave ratio or VSWR (sometimes pronounced ""viswar""). For example, the VSWR value 1.2:1 denotes an AC voltage due to standing waves along the transmission line reaching a peak value 1.2 times that of the minimum AC voltage along that line. The SWR can as well be defined as the ratio of the maximum amplitude to minimum amplitude of the transmission line's currents, electric field strength, or the magnetic field strength. Neglecting transmission line loss, these ratios are identical.The power standing wave ratio (PSWR) is defined as the square of the VSWR, however this terminology has no physical relation to actual powers involved in transmission.The SWR can be measured with an instrument called an SWR meter. Since SWR is defined relative to the transmission line's characteristic impedance, the SWR meter must be constructed for that impedance; in practice most transmission lines used in these applications are coaxial cables with an impedance of either 50 or 75 ohms. Checking the SWR is a standard procedure in a radio station, for instance, to verify impedance matching of the antenna to the transmission line (and transmitter). Unlike connecting an impedance analyzer (or ""impedance bridge"") directly to the antenna (or other load), the SWR does not measure the actual impedance of the load, but quantifies the magnitude of the impedance mismatch just performing a measurement on the transmitter side of the transmission line.