## Root-Mean Square Definition in Physics

**Definition:** The root mean square (rms) current in physics is the steady or direct current (DC) which produces the same heating effect per second in a given resistor. The symbol for root mean square value is **I _{rms}**.

The **formula for calculating the root mean square value is I _{rms }= I_{0} / √2 = 0.71 x I_{0}**

Where I_{0} is the peak current.

**Peak Current (I _{0}):** Is the peak value of an alternating current (AC) circuit which is the maximum value of the current recorded in an AC cycle. The formula for peak current is:

**I**

_{0}= √2 x I_{rms}= 1.414 x I_{rms}We also have **Root mean square voltage** (V_{rms}). The formula for root mean square voltage is **V _{rms} = V_{0} / √2 = 0.71 x V_{0}**. Where V

_{0}is the peak voltage value.

It is also important to know that we can represent alternating current and alternating voltages by:

**Alternating current, I = I _{0}sin2πft = I_{0}sinωt**

And the **alternating voltage, V = V _{0}sin2πt = V_{0}sinωt **

From the above symbols, we have:

I = Instantaneous current (si unit = Ampere(A))

I_{0} = Peak or maximum current (si unit = Ampere(A))

f = frequency (si unit = Heartz (Hz))

t = time (s) (si unit = seconds (s))

ωt = θ = phase angle of current or voltage

V = voltage

V_{0} = Peak voltage

## Root Mean Square Solved Problems

Here are a few problems to help you understand how to calculate rms in physics:

### Problem 1

The current, I in an a.c circuit is given by the equation: **I = 30sin100πt **where t is the time in seconds. Deduce the following from this equation:

a. Frequency of the current

b. Peak value of the current

c. rms value of the current

#### Solution

**Data:** The information from the question

We were given **I = 30sin100πt**, and by comparing it with **I = I _{0}sin2πft** we will obtain

a. Frequency, f as **100πt** = **2πft**

t and π will cancel each other leaving us with

f = 100 / 2 = 50 Hz

**Therefore, the frequency is 50 Hz **

b. The peak current is already available from the equation **I = 30sin100πt**.

**Therefore: I _{0} = 20 A**

c. We can calculate the rms value by applying the formula: **I _{rms }= I_{0} / √2 = 0.71 x I_{0}**

Thus, **I _{rms } = 0.71 x I_{0}** = 0.71 x 30 = 21.3 A

**Therefore, the rms value is 21.3 A**.

### Problem 2

Calculate the peak voltage of a main supply of rms value 220V.

#### Solution

We have V_{rms} = 220 volts

**Applying the formula: V _{0} = V_{rms} x 1.414 = 220 x 1.414 = 311.08 Volts**

### Problem 3

The current through a resistor in an a.c circuit is given as **2sinωt**. Determine the d.c equivalent of the current.

#### Solution

We have **I** **= I _{0}sinωt. **Therefore, the peak current is

**I**

_{0}= 2AThe direct current (DC) equivalent of a.c is the root mean square current I_{rms}

Therefore, I_{rms }= I_{0} / √2 = 2 / √2

We will now multiply both the numerator and denominator by √2 to obtain

I_{rms} = (2/√2) x (√2/√2) = (2√2) / (√2×2) = (2√2) / 2 = √2 Ampere

### Problem 4

If an a.c is represented by I = I_{0}sinωt. Calculate the instantaneous value of such a current, if in a circuit it has rms value of 15.0 A when it is phase angle is 30^{0}.

#### Solution

The equation available to us is I = I_{0}sinωt. And ωt = 30^{0}, I_{rms} = 15.0 A

To find I_{0} will use the formula: I_{0} = 1.414 x I_{rms} = 1.414 x 15 = 21.21 A

**Therefore, we will now apply the equation: I = I _{0}sinωt = 21.21 x sin30^{0} = 10.605 Ampere**

### Problem 5

In an a.c circuit the peak of the potential difference 180V. What is the instantaneous potential difference when it has reached (1/8)th of a cycle?

#### Solution

We have a formula for instantaneous potential difference as: V = V_{0}sinωt

where V = ?, V_{0} = 180V,

ωt = θ = phase angle of the voltage = (1/8)th of a cycle = (1/8) x 360^{0} = 45^{0} [Remember that 1 cycle = 360^{0}]

sin45^{0} =

Thus, to find V, we will substitute our data into the formula:

**V = V _{0}sinωt = 180 x sin45^{0} = 180 x √2/2 = 90√2 = 127 Volts.**

### Problem 6

The instantaneous value of the induced e.m.f as a function of time is ε = ε_{0}sinωt where ε_{0} = is the peak value of the e.m.f. The instantaneous value of the e.m.f, one-quarter of the period is

#### Solution

ωt = θ = phase angle of the e.m.f

Therefore, ωt = θ = one-quarter of the period is = (1/4) x 360^{0} = 90^{0}

By applying the equation available to us from the question, we will obtain:

ε = ε_{0}sinωt = ε_{0}sin90^{0} = ε_{0}

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