An alternating voltage V(t) = 220 sin 100πt volt is applied to a purely resistive load of 50 Ω. The

Q: Why didn't we use the $50 \,\Omega$ resistance or the $220 \text{ V}$ peak voltage?

Because the question asks for the *time* it takes to reach a certain *fraction* of the peak value. The amplitude ($I_0$) cancels out when you set up the ratio $\frac{I_0}{2} = I_0 \sin(\omega t)$. The time depends solely on the frequency of the wave.

Q: Why is the time from $0$ to half peak different from half peak to peak?

A sine wave is not a straight line! It rises very steeply near $0$, so it reaches half its peak relatively quickly (in $\frac{1}{600}$ s). As it approaches the peak, it flattens out, taking twice as long ($\frac{2}{600}$ s) to cover the remaining half of the amplitude.

Q: What if the circuit had an inductor instead of a resistor?

If it were purely inductive, the current would lag the voltage by $90^\circ$. However, the *duration* for the current itself to rise from its own half-peak to its own peak would remain exactly the same ($3.33 \text{ ms}$), because the frequency $\omega$ hasn't changed.

Q: Why did we take $\frac{\pi}{6}$ and not $\frac{5\pi}{6}$ for the half-peak time?

The angle $\frac{\pi}{6}$ represents the *first* time the current reaches half its peak while rising from zero. The angle $\frac{5\pi}{6}$ represents the time it hits half-peak while *falling* back down from the peak.

Q: How do we quickly convert seconds to milliseconds?

The prefix "milli" means one-thousandth ($10^{-3}$). So, to convert a value from seconds to milliseconds, you simply multiply it by $1000$. For example, $\frac{1}{300} \text{ s} = \frac{1000}{300} \text{ ms} = 3.33 \text{ ms}$.

Question

An alternating voltage $V(t) = 220 \sin 100\pi t$ volt is applied to a purely resistive load of $50 \Omega$. The time taken for the current to rise from half of the peak value to the peak value is

Options:

(A) 5 ms

(B) 2.2 ms

(C) 7.2 ms

(D) 3.3 ms

Sigma Percentile · Accepted Answer

### The Setup: A Purely Resistive Dance

Imagine a simple circuit where an alternating voltage is applied across a resistor. The voltage is given by the equation:
$$V(t) = 220 \sin(100\pi t)$$

Because the circuit is purely resistive, the current dances perfectly in sync with the voltage. There is no lagging or leading. The current will also be a sine wave, starting from zero, reaching a peak, and oscillating. We can write the current as:
$$I(t) = I_0 \sin(100\pi t)$$

Our mission is to find the time it takes for this current to rise from exactly half of its peak value to its full peak value. Let's call the time at half peak $t_1$, and the time at the peak $t_2$. We need to find the difference, $\Delta t = t_2 - t_1$.

### Finding the Milestones

Let's find $t_1$ first. At $t_1$, the current is exactly half of its maximum value, so $I(t_1) = \frac{I_0}{2}$. Substituting this into our current equation, we get:
$$I_0 \sin(100\pi t_1) = \frac{I_0}{2}$$

The $I_0$ cancels out beautifully, leaving us with:
$$\sin(100\pi t_1) = \frac{1}{2}$$

We know from basic trigonometry that the sine function equals $\frac{1}{2}$ at an angle of $\frac{\pi}{6}$ radians. Therefore:
$$100\pi t_1 = \frac{\pi}{6}$$
$$t_1 = \frac{1}{600} 	ext{ s}$$

Now for the peak time, $t_2$. At the peak, the current is simply $I_0$. This means:
$$I_0 \sin(100\pi t_2) = I_0$$
$$\sin(100\pi t_2) = 1$$

The sine function reaches its maximum value of $1$ at $\frac{\pi}{2}$ radians. So:
$$100\pi t_2 = \frac{\pi}{2}$$
$$t_2 = \frac{1}{200} 	ext{ s}$$

### The Final Stretch

We have our two crucial timestamps! The time taken to rise from half peak to peak is simply $t_2 - t_1$:
$$\Delta t = \frac{1}{200} - \frac{1}{600}$$

Taking a common denominator of $600$, we get:
$$\Delta t = \frac{3 - 1}{600} = \frac{2}{600} = \frac{1}{300} 	ext{ s}$$

The options provided are in milliseconds. To convert seconds to milliseconds, we multiply by $1000$:
$$\Delta t = \frac{1}{300} 	imes 1000 	ext{ ms} = 3.33 	ext{ ms}$$

**And that's our final answer!**

### The Grand Takeaway

Notice something fascinating here? The resistance value of $50 \,\Omega$ and the peak voltage of $220 	ext{ V}$ didn't even matter! The time taken depends entirely on the angular frequency ($\omega = 100\pi$) of the AC source.

If we doubled the frequency, this time would be halved. Always look for these hidden symmetries and independent variables in physics problems—they often reveal the true nature of the phenomenon!

Animated Solution for Physics - Electromagnetic Induction: An alternating voltage $V (t) = 220 sin 100 π t$ volt is applied to a purely resistive load of $50Ω$ . The time taken for the current to rise from half of the peak value to the peak value is

Select Answer:

Visualized Solution

Visualizing the Circuit

Current Equation

Defining the Goal

Setting up $t_{1}$

Solving for $t_{1}$

Setting up $t_{2}$

Solving for $t_{2}$

Calculating $Δ t$

Final Answer

The Way Forward

The Sigma Insight: Alternating Current (AC) and Voltage

The Setup

Finding the Milestones

The Final Stretch

The Grand Takeaway

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