Question

Q.
Two bulbs of ratings 20v-200w and 20v-50w are connected in series with 40w power supply find total current and power consumed.​

Answer 1

Correct option is

B

40W

Resistance of 100Wbulb=200

2

/100=400ohms

Resistance of 60Wbulb=200

2

/60=666.67ohms

Resistance of 40Wbulb=200

2

/40=1000ohms

Therefore, total resistance in series = (400+666.67+1000)=2066.67ohms

Current in the circuit =200/2066.67=0.0967A

Therefore, actual power consumed by “40W”bulb=0.0967

2

x1000=9.35W (much lesser than any of the original)

The 40W bulb will grow the brightest as the current is constant in all three and it has the maximum resistance. But it would consume much less than 40W as the bulbs are connected in series, and voltage would be divided across all three filaments depending upon resistances.

Answer 2

Here's your answer... hope this helps you;)

Answer Explanation:

Power depends on three factors :

1.Voltage

2.Current

3.Resistance

Of these three, the first two are liable to change according to a given circuit, but the resistance of a device is fairly constant.

Let's take a 200 volts - 100 watts bulb and calculate the resistance of its filament :

We know,

P = $\frac{V^{2} }{R}$

i.e.

R =$\frac{V^{2} }{P}$

When V=200 volts and P = 100 watts,

R = (200*200)/100 = 400 ohms.

The reason why we calculated out the resistances is because R does not change with the applied voltage. It may be considered as a constant at a given temperature and pressure.

Now, let's connect the two bulbs in series :

Since now the two bulbs are in series, the total resistance offered by them is :

R' = 400 Ω + 400 Ω = 800 Ω

Let them be powered by a 200 volts DC source.

Now the current that flows through either of them is :

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

 =  $\frac{200}{800}$

 =0.25 A

Since the bV=IR  

Since the bulbs are in series, the same amount of current flows through each of them.

Now note that the potential across the each bulb in this case is not 200 volts. Instead, it is the voltage across the combination of bulbs.

Since the two bulbs are identical, the voltage across each bulb is :

V=IR

V=0.25∗400

 V=100 volts

In this case, the power at which each bulb is working is given by :

P  =  V∗  I

If we substitute V ( = I*R) from Ohm's law, we get

P=$I^{2}$∗R

P=(0.25∗0.25)∗400  

P = 25 watts

Thus, on connecting the bulbs in series, the power they consume decreases significantly.

This happened because the bulb which was originally rated at 100W at 200V has now been connected to a voltage supply of 100V.