In a 100% efficient step-up transformer, the output voltage is $5$ times greater than the input voltage. How does the output current $I_s$ compare to the input current $I_p$ ?

An ideal step-down transformer reduces voltage from $240\ \text{V}$ AC to $12\ \text{V}$ AC. How does the current in the secondary coil ( $I_s$ ) compare to the current in the primary coil ()?

During one complete $360^{\circ}$ rotation of a single coil within a simple AC generator, how many times does the induced voltage reach its peak magnitude (either positive or negative)?

MYP MYP Physics Topic AC & DC Questionbank

Practice MYP MYP Physics Topic AC & DC with authentic exam-style questions for both SL and HL students. This question bank focuses on the exact syllabus content for AC & DC and mirrors Paper 1, 2 style where relevant.

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The battery connected to the generator reverses its polarity

The magnetic field switches its North and South poles every half turn

The coil stops moving momentarily at every half turn

Each side of the coil reverses its direction of motion relative to the magnetic field

$I_s$ is $5$ times $I_p$

$I_s$ is $25$ times $I_p$

$I_s$ is equal to $I_p$

$I_s$ is $1/5$ of $I_p$

The secondary current is zero because it is a step-down transformer

The secondary current is lower than the primary current

The secondary current is exactly equal to the primary current

The secondary current is higher than the primary current