The electrical energy (in kWh) consumed in operating a bulb of 40 W for 5 hours a day in a month of 30 days is:
First, calculate the total energy consumed by the bulb in a day: $$E_{\text{day}} = P \times t = 40 \, \text{W} \times 5 \, \text{hours} = 200 \, \text{Wh}$$
Next, calculate the total energy consumed in a month of 30 days: $$E_{\text{month}} = E_{\text{day}} \times \text{Number of days} = 200 \, \text{Wh} \times 30 = 6000 \, \text{Wh}$$
Finally, convert the energy from Wh to kWh: $$E_{\text{kWh}} = \frac{6000 \, \text{Wh}}{1000} = 6 \, \text{kWh}$$
Thus, the electrical energy consumed is 6 kWh.
An ideal low pass filter has frequency response given by \[ H(j\omega) = \begin{cases} 1, & |\omega| \leq 200\pi \\ 0, & \text{otherwise} \end{cases} \] Let \( h(t) \) be its time domain representation. Then h(0) = _________ (round off to the nearest integer).
Using shunt capacitors, the power factor of a 3-phase, 4 kV induction motor (drawing 390 kVA at 0.77 pf lag) is to be corrected to 0.85 pf lag. The line current of the capacitor bank, in A, is __________ (round off to one decimal place).
In the Wheatstone bridge shown below, the sensitivity of the bridge in terms of change in balancing voltage \( E \) for unit change in the resistance \( R \), in V/Ω, is __________ (round off to two decimal places).
An air filled cylindrical capacitor (capacitance \( C_0 \)) of length \( L \), with \( a \) and \( b \) as its inner and outer radii, respectively, consists of two coaxial conducting surfaces. Its cross-sectional view is shown in Fig. (i). In order to increase the capacitance, a dielectric material of relative permittivity \( \varepsilon_r \) is inserted inside 50% of the annular region as shown in Fig. (ii). The value of \( \varepsilon_r \) for which the capacitance of the capacitor in Fig. (ii), becomes \( 5C_0 \) is
An ideal sinusoidal voltage source \( v(t) = 230\sqrt{2} \sin(2\pi \times 50t) \, \text{V} \) feeds an ideal inductor \( L \) through an ideal SCR with firing angle \( \alpha = 0^\circ \). If \( L = 100 \, \text{mH} \), then the peak of the inductor current, in ampere, is closest to: