SPH 3U Lesson 6
... f) Try reversing the current (Switch the alligator clips connected to the power supply). What happens? Oersted’s Principle: When a charge is moved through a straight conductor, a circular magnetic field is created around the conductor. Right-Hand Rule (RHR) for a straight conductor: If you hold the ...
... f) Try reversing the current (Switch the alligator clips connected to the power supply). What happens? Oersted’s Principle: When a charge is moved through a straight conductor, a circular magnetic field is created around the conductor. Right-Hand Rule (RHR) for a straight conductor: If you hold the ...
Video Transcript - Rose
... If we set I2 at zero, we can find z11. [math equation] The voltage that crosses the first port is V1. Let’s look at the top node. We can label the current flow through the two resistors as I3 and I4. Based on the Kirchoff Current Law, we know that I1 should be equal to the sum of I3 and I4. [math eq ...
... If we set I2 at zero, we can find z11. [math equation] The voltage that crosses the first port is V1. Let’s look at the top node. We can label the current flow through the two resistors as I3 and I4. Based on the Kirchoff Current Law, we know that I1 should be equal to the sum of I3 and I4. [math eq ...
Lab 1
... resistor and the oscilloscope from the MOSFET drain. With GS at 5 V, measure DS with the multi-meter. This resistance is DS ON for GS = 5 V; note that the multi-meter supplies a very small voltage when used as an ohmmeter. Construct the NOT gate from Figure 2 and connect its input to a switch and 1 ...
... resistor and the oscilloscope from the MOSFET drain. With GS at 5 V, measure DS with the multi-meter. This resistance is DS ON for GS = 5 V; note that the multi-meter supplies a very small voltage when used as an ohmmeter. Construct the NOT gate from Figure 2 and connect its input to a switch and 1 ...
Module 4, Lecture 4: Electric Current and Resistance Electric
... Module 4, Lecture 4: Electric Current and Resistance Current – any motion of ___________ from one region to another. ...
... Module 4, Lecture 4: Electric Current and Resistance Current – any motion of ___________ from one region to another. ...
Dec
... parallel are placed in series with a regulating resistance of 5 ohms and connected across 200 volt mains. Calculate the magnitude and direction of the current in each battery and the total current taken from the supply mains. 2. A D.C. series motor runs at 17 rev/s when taking 20 amperes at 200 volt ...
... parallel are placed in series with a regulating resistance of 5 ohms and connected across 200 volt mains. Calculate the magnitude and direction of the current in each battery and the total current taken from the supply mains. 2. A D.C. series motor runs at 17 rev/s when taking 20 amperes at 200 volt ...
Worksheet for Exploration 31.5
... Worksheet for Exploration 31.5: RL Circuits and Phasors Assume an ideal power supply. The graph shows the voltage as a function of time across the source (red), the resistor (blue), and the inductor (green) (voltage is given in volts and time is given in seconds). Restart. To analyze the currents an ...
... Worksheet for Exploration 31.5: RL Circuits and Phasors Assume an ideal power supply. The graph shows the voltage as a function of time across the source (red), the resistor (blue), and the inductor (green) (voltage is given in volts and time is given in seconds). Restart. To analyze the currents an ...
Example 15 4 kΩ 1 kΩ 2 kΩ
... amps. Let us identify all the nodes in this circuit. This is a node since the 4V voltage source and 1kΩ resistor meet here. We will skip it because we know the node voltage is exactly just 4V. This node is a connection of 3 branches and the node voltage is labeled as v1 . The input node of the op am ...
... amps. Let us identify all the nodes in this circuit. This is a node since the 4V voltage source and 1kΩ resistor meet here. We will skip it because we know the node voltage is exactly just 4V. This node is a connection of 3 branches and the node voltage is labeled as v1 . The input node of the op am ...
ELTK1200 Assignment #6 Solutions
... XL = XC j resistive circuit In the notes, we have dealt with XL > XC and XC > XL. At a specific frequency XL = XC, the inductive and capacitive reactances cancel each other out. This frequency is called resonant frequency and it has a special importance in the electrical world. For us, at this point ...
... XL = XC j resistive circuit In the notes, we have dealt with XL > XC and XC > XL. At a specific frequency XL = XC, the inductive and capacitive reactances cancel each other out. This frequency is called resonant frequency and it has a special importance in the electrical world. For us, at this point ...
Lecture Set 6-Current and Resistance
... This implies a difference in potential since E=DV/d We assume that the difference in potential is small and that it can often be neglected. In this chapter, we will consider this difference and what causes it. ...
... This implies a difference in potential since E=DV/d We assume that the difference in potential is small and that it can often be neglected. In this chapter, we will consider this difference and what causes it. ...
Measuring e/k
... Apparatus: Breadboard, small connecting wires, 2-alligator clips, 2-resistors, and white DMM. Solder-less breadboards are important in electronics. They allow us to make quick circuits, test out ideas before making a more permanent circuit. They often look like the one shown below on the right. Alth ...
... Apparatus: Breadboard, small connecting wires, 2-alligator clips, 2-resistors, and white DMM. Solder-less breadboards are important in electronics. They allow us to make quick circuits, test out ideas before making a more permanent circuit. They often look like the one shown below on the right. Alth ...
solns
... Tau = L/ R = 30 mH / 3 kOhm = 10 microsec 31) For the parallel RLC circuit shown, find alpha, wo, s1, and s2 alpha = 1/(2RC) = 1/(2*1000*1.5x10-6) = 333 wo = sqrt(1/LC) = sqrt(1/(2*1.5x10-6)) = 149 s1 = -333 + sqrt(3332 - 1492) = -35 s2 = -333 – sqrt(3332 - 1492) = -631 ...
... Tau = L/ R = 30 mH / 3 kOhm = 10 microsec 31) For the parallel RLC circuit shown, find alpha, wo, s1, and s2 alpha = 1/(2RC) = 1/(2*1000*1.5x10-6) = 333 wo = sqrt(1/LC) = sqrt(1/(2*1.5x10-6)) = 149 s1 = -333 + sqrt(3332 - 1492) = -35 s2 = -333 – sqrt(3332 - 1492) = -631 ...
BMLR2 Unit 5
... is a closed electrical path for current. 2. A circuit has only one path for current. 3. A has more than one path for current. 4. A circuit, also called combination circuit, is a combination of series circuit and a parallel circuit. ...
... is a closed electrical path for current. 2. A circuit has only one path for current. 3. A has more than one path for current. 4. A circuit, also called combination circuit, is a combination of series circuit and a parallel circuit. ...
TRIAC
TRIAC, from triode for alternating current, is a genericized tradename for an electronic component that can conduct current in either direction when it is triggered (turned on), and is formally called a bidirectional triode thyristor or bilateral triode thyristor.TRIACs are a subset of thyristors and are closely related to silicon controlled rectifiers (SCR). However, unlike SCRs, which are unidirectional devices (that is, they can conduct current only in one direction), TRIACs are bidirectional and so allow current in either direction. Another difference from SCRs is that TRIAC current can be enabled by either a positive or negative current applied to its gate electrode, whereas SCRs can be triggered only by positive current into the gate. To create a triggering current, a positive or negative voltage has to be applied to the gate with respect to the MT1 terminal (otherwise known as A1).Once triggered, the device continues to conduct until the current drops below a certain threshold called the holding current.The bidirectionality makes TRIACs very convenient switches for alternating-current (AC) circuits, also allowing them to control very large power flows with milliampere-scale gate currents. In addition, applying a trigger pulse at a controlled phase angle in an AC cycle allows control of the percentage of current that flows through the TRIAC to the load (phase control), which is commonly used, for example, in controlling the speed of low-power induction motors, in dimming lamps, and in controlling AC heating resistors.