Question

the student then uses the circuits to investigate the safety devices used in electric circuits. the diagram below shows three different fuses with their current ratings show
fuse x 6a fuse y 3a fuse z 1a
(b) what does the current rating of a fuse indicate?

she adds fuse x into circuit 1 as shown in the diagram below.
diagram of a circuit with battery, ammeter a, bulb a, and fuse x
assume that the fuses, x, y and z, have no resistance.
(c) suggest whether the brightness of the bulb in circuit 1 increases, decreases or remains the same when fuse x is added.

(d) the student replaces fuse x with fuse y, then with fuse z. state and explain, for each fuse, whether the bulb lights up.
(i) fuse y

Explanation:

Part (b)

The current rating of a fuse indicates the maximum current that the fuse can safely carry without melting (blowing). If the current in the circuit exceeds this rating, the fuse wire heats up, melts, and breaks the circuit to protect the electrical components from damage due to excessive current (like short - circuits or overloading).

We know from Ohm's law \(I = \frac{V}{R}\) (where \(I\) is current, \(V\) is voltage, and \(R\) is resistance) and the power formula for a bulb \(P=I^{2}R_{bulb}\) (where \(P\) is power and \(R_{bulb}\) is the resistance of the bulb). The problem states that the fuse has no resistance. The total resistance of the circuit before adding the fuse is \(R_{bulb}\), and after adding fuse \(X\) (with no resistance), the total resistance of the circuit \(R_{total}=R_{bulb}+ 0=R_{bulb}\). Since the voltage of the power source \(V\) remains constant, from \(I=\frac{V}{R_{total}}\), the current in the circuit remains the same. And since the power dissipated by the bulb \(P = I^{2}R_{bulb}\) depends on the current and the bulb's resistance (both of which remain unchanged), the brightness of the bulb, which is related to the power it dissipates, remains the same.

First, we need to know the current rating of fuse \(X\) (let's assume from the diagram that fuse \(X\) has a current rating that allows the normal current of the circuit to pass, and fuse \(Y\) has a current rating. Let's assume the current in the circuit (when the bulb is connected) is \(I_{circuit}\). If the current rating of fuse \(Y\) is greater than or equal to \(I_{circuit}\), then the fuse will not blow. Since the fuse has no resistance, the circuit remains complete, and current will flow through the bulb, causing it to light up. We assume that the current drawn by the bulb is less than the current rating of fuse \(Y\) (as fuse \(Y\) is being tested after fuse \(X\) which worked). So, the fuse \(Y\) will not blow, the circuit is closed, and current flows through the bulb, so the bulb lights up.

Answer:

The current rating of a fuse indicates the maximum current the fuse can carry without melting (blowing) to protect the circuit.

Part (c)