An AC current is given by I = I1 sin(wt) + I2 cos(wt). A hot wire ammeter will give a reading
(a) sq

Q: Why does a hot wire ammeter measure RMS and not average current?

A hot wire ammeter works on the heating effect of electric current. The heat produced is given by Joule's law, $H = I^2 R t$, which depends on the square of the current. Therefore, the deflection is proportional to the mean square of the current, making it calibrated to read the Root Mean Square (RMS) value.

Q: Why is the average of $\sin \omega t \cos \omega t$ zero?

Using the trigonometric identity $2 \sin \theta \cos \theta = \sin 2\theta$, we can rewrite the term as $\frac{1}{2} \sin 2\omega t$. The average of any pure sine or cosine wave over a full cycle is zero because the positive half-cycle perfectly cancels out the negative half-cycle.

Q: Can we just add the RMS values of the two individual components?

No, RMS values do not add linearly. You cannot say $I_{rms} = I_{rms1} + I_{rms2}$. Instead, if the components are orthogonal (like sine and cosine, or DC and AC), you must add their squares: $I_{rms} = \sqrt{I_{rms1}^2 + I_{rms2}^2}$.

Q: What if the frequencies of the two terms were different, like $\sin \omega_1 t$ and $\cos \omega_2 t$?

As long as the frequencies are different, the time average of their cross product $\langle \sin \omega_1 t \cos \omega_2 t \rangle$ over a sufficiently long time (or a common period) will still be zero. Thus, the final RMS formula remains the same: the square root of the sum of the squares of their individual RMS values.

Q: Is this formula applicable to any periodic waveform?

The fundamental definition $I_{rms} = \sqrt{\frac{1}{T} \int_0^T I^2 dt}$ applies to any periodic waveform, whether it is a square wave, a triangular wave, or a complex sum of sines. You just need to integrate the square of the specific function over its period.

Question

An AC current is given by $I = I_1 \sin \omega t + I_2 \cos \omega t$. A hot wire ammeter will give a reading

Options:

(A) $\sqrt{\frac{I_1^2 - I_2^2}{2}}$

(B) $\sqrt{\frac{I_1^2 + I_2^2}{2}}$

(C) $\frac{I_1 + I_2}{\sqrt{2}}$

(D) $\frac{I_1^2 + I_2^2}{2\sqrt{2}}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are looking at a circuit with a special kind of measuring device: a **hot wire ammeter**. Unlike standard moving coil galvanometers that measure the average current (which would be zero over a full AC cycle), a hot wire ammeter works on the principle of the heating effect of current. Because the heat produced is proportional to the square of the current ($H = I^2 R t$), this device effectively measures the **Root Mean Square (RMS)** value of the alternating current.

Our goal is to find the reading of this ammeter for a given complex current waveform:

$$I = I_1 \sin \omega t + I_2 \cos \omega t$$

### The Master Equation

To find the RMS value, we must follow its mathematical definition strictly. The RMS current is the square root of the mean (time average) of the square of the current over one complete cycle.

$$I_{rms} = \sqrt{\langle I^2 \rangle} = \sqrt{\frac{1}{T} \int_0^T I^2 dt}$$

Let's start by squaring the instantaneous current equation. We substitute the given expression for $I$ and square the entire binomial.

$$I^2 = (I_1 \sin \omega t + I_2 \cos \omega t)^2$$

Don't make a silly mistake here; we need to expand this using the algebraic identity $(a + b)^2 = a^2 + b^2 + 2ab$.

$$I^2 = I_1^2 \sin^2 \omega t + I_2^2 \cos^2 \omega t + 2 I_1 I_2 \sin \omega t \cos \omega t$$

### Time Averaging the Components

Now, we need to find the mean, or time average, of this entire expression over one full cycle. Let's recall some standard results for trigonometric averages over a complete cycle $T = \frac{2\pi}{\omega}$:

1. The average of $\sin^2 \omega t$ is exactly $\frac{1}{2}$.
2. The average of $\cos^2 \omega t$ is exactly $\frac{1}{2}$.
3. The average of the cross term $\sin \omega t \cos \omega t$ is $0$. This is because $2 \sin \omega t \cos \omega t = \sin 2\omega t$, and the average of any pure sine wave over a full cycle is zero.

Let's substitute these average values back into our expanded equation.

$$\langle I^2 \rangle = I_1^2 \left(\frac{1}{2}\right) + I_2^2 \left(\frac{1}{2}\right) + 2 I_1 I_2 (0)$$

This simplifies beautifully. The cross term completely vanishes, leaving us with a clean expression for the mean square current.

$$\langle I^2 \rangle = \frac{I_1^2 + I_2^2}{2}$$

### Final Calculation

Finally, to get the RMS current, which is exactly what the hot wire ammeter reads, we simply take the square root of our mean square value.

$$I_{rms} = \sqrt{\langle I^2 \rangle} = \sqrt{\frac{I_1^2 + I_2^2}{2}}$$

This perfectly matches option (b).

As a thought experiment, what if the current equation had a direct current (DC) component, say $I = I_0 + I_1 \sin \omega t$? In that case, the average of $I_0^2$ would just be $I_0^2$, and the cross term would still average to zero. The RMS value would then become $\sqrt{I_0^2 + \frac{I_1^2}{2}}$. Always break complex waveforms down into their mean squares!

Animated Solution for Physics - Electromagnetic Induction: An AC current is given by $I = I_{1} sin ω t + I_{2} cos ω t$ . A hot wire ammeter will give a reading

Select Answer:

Visualized Solution

Hot Wire Ammeter

RMS Current Formula

Squaring the Current

Expanding the Square

Time Average of Trigonometric Functions

Mean Square Current

Final RMS Value

The Way Forward

The Sigma Insight: Alternating Current (AC) and Voltage

Analyzing the Setup

The Master Equation

Time Averaging the Components

Final Calculation

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Animated Solution for Physics - Electromagnetic Induction: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } An AC current is given by I=I1​sinωt+I2​cosωt. A hot wire ammeter will give a reading