For Bulb 1
Power = 60 W
Voltage = 230 V
R = V2/P
R = (230 × 230)/60
R = 881.66 Ω
For Bulb 2
Power = 230 W
Voltage = 120 V
R = V2/P
R = (230 × 230)/100
R = 529 Ω
In a parallel connection, the voltage across each element is the same. So when a 60W bulb and 1o0 W bulb are connected in parallel, the voltage across them will be the same i.e 230 V in the given case. To find which bulb will glow brighter we need to find the power dissipation across each of them. From the relation
P = V2/R
Since the voltage is the same we can say that power dissipation will be higher for the bulb with lower resistance i.e. 100W bulb, or the bulb with the higher rating will glow brighter.
In a series connection, the current flowing across each element is the same. So when the 60W bulb and 100W bulb are connected in series, the same current will flow through them. To find which bulb will glow brighter we need to find the power dissipation across each of them. From the relation
P = I2R
The hexadecimal number system is also called base-16, a numeration system in which all numbers are represented using the symbols 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, and F only. The system is base of 16. The hexadecimal numbers are used to represent binary numbers because of ease of conversion and compactness.
In the above option, the wood has the lowest calorific value. Its value is around 17000 – 20000 kilojoule/kg.
The calorific value of LPG is 55000 kilojoule/kg.
The calorific value of Kerosene oil is 45000 kilojoule/kg.
The calorific value of hydrogen oil is 150000 kilojoule/kg.
Hysteresis Loss = Kh × BM1.67 × f × v watts
Kh = Hysteresis constant depends upon the material
Bm = Maximum flux density
f = frequency
v = Volume of the core
Hence the hysteresis loss does not depend upon the ambient temperature.
Synchronous speed NS = 120f/P
P = 12 & F = 60 then Ns = ?
NS = 120f/P = (120 × 60)/12 = 600 RPM