# Basic Electrical Engineering (3110005)

BE | Semester-1   Winter-2019 | 11-01-2020

## Q1) (b)

#### A 100V, 60Watt bulb is to be operated from a 220V supply. What is the resistance to be connected in series with the bulb to glow normally ?

Given Data

Supply Voltage,
Bulb rated voltage,
Bulb rated power,

External resistance to be connected in series with bulb,

Calculations
The relation between Power, Voltage and Resistance for the bulb is,

Hence, the internal resistance of the bulb is,

Therefore, the rated current to be passed through the bulb is,

• The total supply voltage is 220 V AC, supplied by the single-phase AC voltage source. This voltage will drop across the bulb and series connected resistance.
• Hence, the Supply voltage = Voltage drop across the bulb + Voltage drop across the series resistance
220 = 100 + Voltage drop across the series resistance
Voltage drop across the series resistance = 220 – 100 = 110 V.
• Now, the voltage across the series resistance is 110 V and the current passing through the series resistance is 0.60 A.
• Therefore, from Ohm’s law, the value of series resistance R is