Two electric bulbs marked 25W – 220V and 100W – 220V are connected in series to a 440V supply. Which of the bulbs will fuse?

###### Solution.

First, let’s calculate the resistance of the bulbs.

We could use the following formula

R = V^2/P ,

where R is the resistance in ohms, P is the electric power in watt and V is the voltage in volts.

Then,

R_{1} = 220^2/25 = 1936 Ω

R_{2} = 220^2/100 = 484 Ω

The total resistance of the series circuit is found by simply adding up the resistance values of the individual resistors.

Hence,

R_{total} = R_{1} + R_{2} = 1936 + 484 = 2420 Ω

The total current in the circuit is given by the following formula

I = V_{total}/R_{total}

so,

I = 440/2420 ≈ 0.18 A

The maximum current that can be allowed to pass through a bulb is given by the formula

I = P/V

Hence,

For the first bulb: I_{1} = 25/220 ≈ 0.11 A

For the second bulb: I_{2} = 100/220 ≈ 0.45 A

I > I_{1}

The total current is bigger than the allowed value on the first bulb, so the 25W bulb will fuse.

Source: sawaal

Answer uncompleted. What’s the status of second bulb (100w).

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