Question

Two bulbs of 25W and 100W rated at 200 volts are connected across a 440V in series
a) The bulbs that will glow brighter.
b) Find the Equivalent resistance
c) What happen if both are connected in parallel. ​

Answer 1

Answer:

two bulbs of 25 W and 100 W rated at 200 volts are connected in series across a 440 volts supply, then. Voltage across 25 W bulb is greater than specified voltage, hence it will fuse. Hence, (B) is correct.

Answer 2

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25 Watt bulb will fuse first.

Resistance of each bulb can be calculated from power ratings and voltage ratings.

Power = V*I = V*(V/R) =V^2/R. Therefore:

R = V^2/Power.

Resistance of 25 Watt bulb:

R25 = 220*220/25 = =1936 Ohms

Resistance of 100 Watt bulb:

R100 = 220*220/100 = =484 Ohms

When they are connected in series with supply of 440 V, resultant current will be:

440 V/ (1936 +484) Ohms = 0.1819 Amperes.

Voltage drop across 25 W bulb will be:

V25 =Current * R25

= 0.1818*1936 = 352 Volts. Because of the excessive voltage (much higher than rated voltage of 220 V), the bulb will fuse in short time, in less than 1-2 seconds.

Voltage across 100 Watt bulb can be calculated By subtracting 352 from 440 V, which comes to be 88 Volts. To cross check, let us calculate it by multiplication of current and resistance:

Voltage drop across 100 W bulb will be

V100 = Current * R100

= 0.1818*484 = 88 Volts.

100 Watt bulb will not fuse.