Three bulbs of 40w,60w and 100w are arranged on series with 220V, which bulb has minimum resistance? a)100w b)40w c)60w d)equal in all?

Question

Answer

We know that :-->
Power (P) = V^2/R ... so , R = V^2/P
and , here all bulb are connected in series .... so, voltage remaining same i.e 220 v
Now , for power of 40 w bulb ... R = (220×220)/40 =1210 ohm
And , for power of 60 w bulb ... R = (220×220)/60 = 806.66 ohm
And , for power of 100 w bulb .... R = (220×220)/100 = 484 ohm
So, here power of 100 w bulb has minimum resistance

So, option (a)... is correct

$$Hope it's help... $$

Answer

Answer

Explanation:
When bulbs are connected in series, the same current flows through each bulb. Hence, the bulb with the highest resistance will drop the maximum voltage across it, while the bulb with the lowest resistance will drop the minimum voltage across it. Therefore, the bulb with the minimum resistance will be the one that has the minimum voltage drop across it.

Calculation of Resistance:
Resistance can be calculated using Ohm's law, which states that resistance (R) is equal to voltage (V) divided by current (I).
R = V/I

Calculation of Current:
Total resistance in the circuit can be calculated by adding the resistance of each bulb.
Rtotal = R1 + R2 + R3

The current in the circuit can be calculated using Ohm's law.
I = V/Rtotal

Calculation of Voltage:
The voltage across each bulb can be calculated using Ohm's law.
V1 = IR1
V2 = IR2
V3 = IR3

Determination of Minimum Resistance:
The bulb with the minimum resistance will have the minimum voltage drop across it.
Therefore, we need to find the bulb with the lowest voltage drop.

Calculations:

Given: V = 220V, P1 = 40W, P2 = 60W, P3 = 100W

Using the formula P = VI, we can calculate the current in each bulb.

I1 = P1/V = 40/220 = 0.1818 A
I2 = P2/V = 60/220 = 0.2727 A
I3 = P3/V = 100/220 = 0.4545 A

Using Ohm's law, we can calculate the resistance of each bulb.

R1 = V/I1 = 220/0.1818 = 1207.69 Ω
R2 = V/I2 = 220/0.2727 = 806.45 Ω
R3 = V/I3 = 220/0.4545 = 484.85 Ω

The total resistance of the circuit is:
Rtotal = R1 + R2 + R3 = 1207.69 + 806.45 + 484.85 = 2499.99 Ω

The current in the circuit is:
I = V/Rtotal = 220/2499.99 = 0.088 A

The voltage drop across each bulb is:
V1 = IR1 = 0.088 x 1207.69 = 106.15 V
V2 = IR2 = 0.088 x 806.45 = 71.05 V
V3 = IR3 = 0.088 x 484.85 = 42.61 V

Therefore, the bulb with the minimum voltage drop across it (i.e. the minimum resistance) is the bulb with the highest power rating, which is the 100W bulb.

Hence, the answer is option (a) 100W.

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