Two electric bulbs marked 25W – 220V and 100W – 220V are connected in series to a 440V supply. Which of the bulbs will fuse?
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.
R1 = 220^2/25 = 1936 Ω
R2 = 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.
Rtotal = R1 + R2 = 1936 + 484 = 2420 Ω
The total current in the circuit is given by the following formula
I = Vtotal/Rtotal
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
For the first bulb: I1 = 25/220 ≈ 0.11 A
For the second bulb: I2 = 100/220 ≈ 0.45 A
I > I1
The total current is bigger than the allowed value on the first bulb, so the 25W bulb will fuse.