AC Fundamentals NEET 2026: Time to First Current Peak

What is the answer to the AC Fundamentals NEET 2026 question?

Option D is correct: the current first reaches its positive peak after one-quarter of a cycle, not half a cycle. In this AC Fundamentals NEET 2026 question, the starting condition determines which phase to choose.

A sinusoidal current has amplitude 5 A and completes 60 cycles each second. Take the start as its zero crossing towards positive current, and find the earliest time at which it becomes +5 A.

The original choices are:

\frac{1}{120} s

\frac{1}{60} s

\frac{1}{30} s

\frac{1}{240} s

The chapter is Electromagnetic Induction and Alternating Currents. The question bank rates this medium difficulty and lists an expected solving time of 60 seconds, a target rather than a measured student average.

How do you write the current equation from the given frequency?

i = i_{peak} \sin (ω t)

This is the official solution’s zero-phase model: current is zero at the start and positive immediately afterwards. It matches the specified positive-going zero crossing. An unshifted cosine would start at the positive peak, so it would use the wrong starting condition.

The symbols on the left and inside the expression represent instantaneous current, peak current, angular frequency and elapsed time, respectively. Convert frequency into angular frequency before substituting: $f = 60 Hz$ $ω = 2 π f$

ω = 2 π \times 60 = 120 π rad/s

Now substitute the peak current:

i_{peak} = 5 A

$i = 5 \sin (120 π t)$ Here, current is measured in amperes and time in seconds. 60 Hz means 60 complete cycles per second, not an angular frequency of 60 radians per second. Each complete cycle spans: $2 π radians$

How does the first-peak condition give option D?

Set the instantaneous current equal to the positive peak, then choose the earliest positive phase that makes the sine equal to one. Solve the sine equation first; use the quarter-period relation afterwards as a check.

At the required endpoint: $i = 5 A$ $5 = 5 \sin (120 π t)$

Divide both sides by 5: $\sin (120 π t) = 1$

Use the first positive phase satisfying this condition:

1 = \sin (\frac{π}{2})

120 π t = \frac{π}{2}

Therefore:

t = \frac{π / 2}{120 π} = \frac{1}{240} s \approx 4.17 ms

The answer is option D. Later positive peaks also satisfy the sine condition, at phases:

ω t = \frac{π}{2} + 2 π n, n = 0, 1, 2, \dots

The first arrival uses zero for the integer index. Later values add complete cycles and do not answer the question.

The 5 A cancels because amplitude sets the peak’s height. Frequency and the specified starting phase determine when that peak occurs. Changing the amplitude alone does not change the time needed to reach the new positive peak.

Now check the result using the period:

T = \frac{1}{f} = \frac{1}{60} s

t = \frac{T}{4} = \frac{1}{240} s

The quarter-period check applies because the current is sinusoidal and starts at a positive-going zero crossing. Do not use it without checking that starting condition.

Why does option A give zero current instead of the peak?

Option A reaches the next zero crossing, not the positive peak. Taking half the period gives its value, but assigns a half-cycle to the wrong pair of waveform positions. This is an endpoint error, not an arithmetic slip.

The incorrect method calculates the period, then assumes zero-to-peak takes half of it:

T = \frac{1}{60} s

t_{wrong} = \frac{T}{2} = \frac{1}{120} s

For the stated starting condition, the sequence is:

Start:

t = 0, i = 0

Positive peak:

t = \frac{T}{4}, i = + 5 A

Next zero crossing:

t = \frac{T}{2}, i = 0

Negative peak:

t = \frac{3 T}{4}, i = - 5 A

One complete cycle:

t = T, i = 0

Test option A directly in the original equation:

i = 5 \sin (120 π \times \frac{1}{120}) = 5 \sin (π) = 0 A

It cannot be the positive peak. Half a cycle takes this current from one zero crossing to the next, or from a positive peak to a negative peak.

The required endpoint is instantaneous current +5 A, not an RMS value. Keep current magnitude, phase angle and elapsed time separate: identify the required current, then its phase, then its time.

How do you choose the phase in two related AC questions?

Divide the required current by the amplitude, then choose the earliest allowed phase. These are original related practice questions, not additional verified NEET PYQs. Both use the same current equation and positive-going zero crossing: $i = 5 \sin (120 π t)$

When does the current first reach 2.5 A?

The first arrival is approximately 1.39 ms after the positive-going zero crossing. Set the instantaneous current to half the amplitude: $2.5 = 5 \sin (120 π t)$

\sin (120 π t) = \frac{1}{2}

The first phase is:

120 π t = \frac{π}{6}

t = \frac{π / 6}{120 π} = \frac{1}{720} s \approx 1.39 ms

Half the peak current does not mean half the zero-to-peak time. Sine is not a linear function of time; use the phase corresponding to the required fraction.

When does the current first reach negative 5 A?

The first negative peak occurs 12.5 ms after the same start. Retain the negative sign when setting the endpoint: $- 5 = 5 \sin (120 π t)$ $\sin (120 π t) = - 1$

The first positive phase satisfying this condition is:

120 π t = \frac{3 π}{2}

t = \frac{3 π / 2}{120 π} = \frac{1}{80} s = 12.5 ms

A negative phase describes a time before the chosen start. The question asks for the first arrival afterwards.

For your next question, write the sine equation, express the required current as a fraction of the amplitude, choose the earliest allowed phase, then divide by angular frequency.

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For a worked example of the same idea, see NEET Question Selection: What to Attempt, Skip and Revisit.

Frequently asked questions

When does a 60 Hz AC current first reach its positive peak?

For a sinusoidal current starting at a positive-going zero crossing, the first positive peak occurs at T/4 = 1/(4f). At 60 Hz, this gives 1/240 s, approximately 4.17 ms, so option D is correct.

Why is the time from zero current to peak current not T/2?

Starting at a positive-going zero crossing, a sinusoidal current reaches its positive peak at phase π/2, which takes T/4. At T/2, the phase is π and the current is zero again. For the stated 60 Hz current, option A, 1/120 s, therefore gives zero current rather than the peak.

How do I write the current equation for a 5 A peak current at 60 Hz?

Convert frequency to angular frequency using ω = 2πf = 120π rad/s. With a positive-going zero crossing at t = 0, the equation is i = 5 sin(120πt), with current in amperes and time in seconds. An unshifted cosine would start at the positive peak and would not match this starting condition.

Does the amplitude affect the time taken to reach the AC peak?

For a sinusoidal current with fixed frequency and starting phase, changing the amplitude alone does not change the time to reach its positive peak. Setting i equal to the peak amplitude makes the amplitude cancel from the sine equation. Frequency and starting phase determine the peak time.