The self-inductance is given as L = μN2A/I L ∝ N2 where N is the number of turns of the solenoid A is the area of each turn of the coil l is the length of the solenoid and μ is the permeability constant L1/L2 = N21/N22 0.5/0.5 = N21/N22 N1/N2 = 1
The average value of the sine wave over one complete cycle is actually zero. Hence, for a sine wave, the average value is defined over half the period. The average value expressed in terms of peak value is given by Average value = 0.637 × peak value
Total resistance in the given circuit R = (250 + 250)MΩ = 500 MΩ Current I = V/R = 24/(500 × 103) Now the Voltage in the voltmeter = \dfrac{{24}}{{500 \times {{10}^3}}} \times 250 \times {10^3} V = 12 V
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
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 due to infinite Linear current-carrying conductor is B = μoI/2πr B/μo = I/2πr H = (I/2πr)A/m
Admittance (Y) is the reciprocal of the impedance of a circuit. Admittance of an AC circuit is analogous to the conductance of a DC circuit. The unit of Admittance is Simen or MHO Admittance = 1/Z simen Y = Conductance ± J Susceptance Or the Admittance can be written as Y = (G ± J B) Simen Now comparing the above equation by the given equation in the question i.e Y= a + jb ∴ a = G = Conductance
EXAMIANS
