Basic Electrical Engineering (3110005)

Question-1b

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

Q1) (b)

Given Data

Supply Voltage,

V_{s} = 220 V

Bulb rated voltage,

V_{b} = 100 V

Bulb rated power,

P_{b} = 60 W

Task

External resistance to be connected in series with bulb,

R = ?

Calculations

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

P_{b} = \frac{V_{b}^{2}}{R_{b}}

Hence, the internal resistance of the bulb is,

R_{b} = \frac{V_{b}^{2}}{P_{b}}

\Rightarrow R_{b} = \frac{100^{2}}{60}

\Rightarrow R_{b} = 166.66 Ω

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

P_{b} = I^{2} R_{b}

\Rightarrow I^{2} = \frac{P_{b}}{R_{b}}

\Rightarrow I^{2} = \frac{60}{166.66}

\Rightarrow I^{2} = 0.3600

\Rightarrow I = 0.60 A

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

$R = \frac{V}{I} = \frac{110}{0.60}$

$R = 183.33 Ω$