Two electric bulbs p and q have their resistances in the ratio of 1:2. They are connected in series across a battery.find the ratio of the power dissipation in these bulbs.
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Two electric bulbs P and Q have resistance in the ratio of 1:2, are connected in series. Find the ratio of power dissipation?
Let the resistances of the bulb be 10 Ohm and 20 Ohm respectively.
When connected in series the equivalent resistance = 10 Ohm + 20 Ohm = 30 Ohm
Let the current supply be having a supply voltage of 30 Volts and zero internal resistance.
Then current I through the series combination of the bulbs = 30 V /30 Ohm = 1 amp
The same current flows through both the bulbs of 10 Ohm and 20 Ohm resistance.
Power dissipation (I² R) through 10 Ohm resistor = 1² × 10 Ohm= 10 Watt
Power dissipation through the 20 Ohm resistor = 1² × 20 = 20 Watt.
Resistance of bulbs in series = 1 : 2
Power dissipation ratio = 10Watt : 20 Watt = 1 : 2.
Let the resistances of the bulb be 10 Ohm and 20 Ohm respectively.
When connected in series the equivalent resistance = 10 Ohm + 20 Ohm = 30 Ohm
Let the current supply be having a supply voltage of 30 Volts and zero internal resistance.
Then current I through the series combination of the bulbs = 30 V /30 Ohm = 1 amp
The same current flows through both the bulbs of 10 Ohm and 20 Ohm resistance.
Power dissipation (I² R) through 10 Ohm resistor = 1² × 10 Ohm= 10 Watt
Power dissipation through the 20 Ohm resistor = 1² × 20 = 20 Watt.
Resistance of bulbs in series = 1 : 2
Power dissipation ratio = 10Watt : 20 Watt = 1 : 2.
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