The portable power supply has an alternator connected to a transformer - AQA - GCSE Physics - Question 7 - 2020 - Paper 1

Using the transformer equation:

$\frac{V_p}{V_s} = \frac{N_p}{N_s}$

where:

• $V_p$ = Primary voltage = 1.5 V
• $V_s$ = Secondary voltage = 5.0 V
• $N_p$ = Number of turns on primary = 150
• $N_s$ = Number of turns on secondary

Now, substituting the values:

$\frac{150}{N_s} = \frac{1.5}{5.0}$

Cross-multiplying gives:

$150 \cdot 5.0 = 1.5 \cdot N_s$
$750 = 1.5 \cdot N_s$

Now, divide both sides by 1.5:

$N_s = \frac{750}{1.5} = 500$

Therefore, the number of turns on the secondary coil is 500.

An alternating current is induced because as the handle of the alternator is turned, the coil moves through a magnetic field. This motion causes the coil to cut magnetic field lines, inducing a potential difference across the coil. Thus, as the coil rotates, the direction of the induced current changes every half turn, resulting in an alternating current.

The slip rings provide a continuous, moveable contact between the rotor (coil) and the external circuit, preventing the wires from twisting together, which ensures a stable connection for the induced current.

When the alternator is disconnected from the transformer and lamp, there is no induced current. Without the opposition from the magnetic field produced by the coil, it requires less force to turn the handle, making it easier to operate.