A fuse is a device to avoid the flow of a large current in a circuit. The potential difference across the fuse is negligible. In our circuit, the maximum current allowed through the fuse is 10 A. The light bulbs marked 40 Watt, 110 V. The bulbs connected to a voltage source of 110 V. What is the maximum number of bulbs that can be placed in parallel before the fuse blows?

###### Solution.

The power dissipated by the bulb is given by

P = V^{2}/R,

where V is the voltage on the light bulb and R is the electrical resistance.

We can calculate the resistance of the light bulb

R = V^{2}/P = 110^{2}/40 = 302.5 Ω.

The total resistance of two bulbs is

1/R_{T} = 1/R + 1/R

or

R_{T} = R/2.

Thus, for n light bulbs, we will get

R_{T} = R/n. (1)

The total current in the circuit is given by

I = V/R_{T} (2) (The Ohm’s law).

We substitute the equation (1) into the equation (2) and obtain

I = V*n/R (3)

If the current I is equal or greater than the maximum current I_{max} then the fuse blows.

I ≥ I_{max}

We can rewrite it using equation (3)

V*n/R ≥ I_{max}

or

n ≥ I_{max}*R/V (4)

We substitute the values into the equation (4) and we obtain

n ≥ 10*302.5/110

n ≥ 27.5

Finally, we can see that we are safe if we are connecting 27 light bulbs in parallel and the fuse blows if we will place 28 light bulbs.