According to the Norton theorem, to find the Norton current, first remove the load resistance RL from the network terminals AB. Short circuit the terminals AB as shown in Figure calculate the current ISc or IN through the short circuit. Now resistance of 150Ω will not show any effect in the circuit. So only resistance of 30Ω will be effective. Norton current IN = 360/30 IN = 12 A
From the figure it can be concluded that the voltmeter reads 5 volts as shown in the figure below.
Based on the voltmeter and ammeter readings in the measuring network, determine the value of the resistor R
Here
Current I = 1/2 A = 0.5 A
Voltage V = 5 V
R = V/I = 0.5/5
R = 10Ω
Galvanized steel conductors do not corrode, and possess high resistance. Hence such Wires are used in telecommunications circuits, earth wires, guard wire, stray wire, etc.
Magnetic Field Strength (H) gives the quantitative measure of strongness or weakness of the magnetic field.
H = B/μo
Where
B = Magnetic Flux Density
μo = Vacuum Permeability
Magnetic Field strength at the center of circular loop carrying current I is given by
B = μoI/2r
B/μo = I/2r
H = I/2r
Where r = Radius
Now Given Parameters
Diameter = 1m
Current = 2A
∴ Magnetic field Intensity H = (2 / 2 × 1/2) = 2 A/m
Capacitor start and capacitor run motor: Two capacitors are used for starting, but one of them is cut out when speed reaches 70 percent of the synchronous speed. The capacitor start-and-run motor starts with a high value and a low-value capacitor connected in parallel with each other but in series with the starting winding. This provides a very high starting torque. The centrifugal switch disconnects the high-value capacitor at 80 percent speed, but the lower value capacitor remains in the circuit.
EXAMIANS
