How to measure AC/DC voltage & current using CRO? Explain with diagrams

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Applications of CRO

The CRO is a versatile laboratory instrument. It is used for all types of measurements like AC/DC voltage, AC/DC current, frequency and phase measurements. These methods are given below.

Measurement of AC/DC voltage

Measurement of AC voltage: With this method, we can measure the peak value of AC voltage. Connect unknown AC voltage across y-input of CRO. A following type of waveform will be displayed on the screen.

Measure its peak displacement in centimeters on graticule scale. Note down the position of Volt/div scale in Volts. Then using the formula calculate the peak value V_p –

V_p = vertical displacement × volt/div

and the RMS value of AC voltage is given by: V_rms = V_p / √2 = V_p / 1.414

In the above example, vertical displacement = 3cm and volt/div knob = 5mV. So peak value = 15mV

Measurement of DC voltage: Connect unknown DC voltage to y-input of CRO. A straight horizontal line will be displayed on screen. If it is above x-axis, then the applied voltage is positive, but if it is below x-axis, the voltage is negative.

Measure its vertical displacement in centimeters from x–axis using graticule scale. Note down volt/div reading and calculate the unknown voltage V_dc as follows –

V_dc = vert. displacement × volt/div

In above example, vert. displacement = 2cm and volt/div = 2V. So peak value will be +4V, because the horizontal line is above x-axis.

Exercise

1. What is the peak value of AC voltage, if its r.m.s. value is 10V? (Ans: Vp = 14.14V)
2. What is the value of unknown DC voltage, if a straight line below x-axis is obtained with a displacement of 3cms? The volt/div = 2V.
   (Ans: Vdc = –6V)
3. Calculate peak-to-peak value of unknown AC voltage, if volt/div = 10V and the peak displacement of x-axis is 2.5cms. (Ans: Vp = 50V)
4. Where the straight line will appear on screen, if we connect 0V DC at y-input of CRO? (Ans: ?)

Measurement of AC/DC Current

We cannot measure current directly using CRO so we use indirect method, as shown below.

First pass the unknown AC/DC current through a known value of resistor, say 10Ω. Then connect the voltage drop across it to y-input of CRO. If the unknown current is AC we will get a waveform, but if it is DC, we will get a straight line. There are two possible conditions:

Measurement of AC current: When voltage drop across 10Ω resistor is connected to y-input, we will get a waveform on the screen. Measure its peak displacement in centimeters on graticule scale. Note down the position of Volt/div scale in Volts.

Then using the formula calculate the peak value V_p –

V_p = vertical displacement × volt/div

Now we can calculate the value of unknown current as –

Unknown current = (vertical displacement × volt/div) / 10Ω

= (2.3 × 5) / 10Ω = 1.15A

Measurement of DC current: Connect the voltage drop across 10Ω resistor to y-input. Note down the position of Volt/div scale. Measure the displacement of the line from x-axis.

Then using the formula calculate the peak value V_dc = vert. displacement × volt/div

Now we can calculate the value of unknown current as –

Unknown current = (vertical displacement × volt/div) / 10Ω

= (2 × 10) / 10Ω = 2A

Since the line is below x-axis, the current is negative i.e. unknown current = – 2A

Exercise

1. An unknown current is passed through a 1Ω resistor and the voltage drop is connected to y-input of CRO. Following waveform is observed on the screen. Calculate the value of unknown current and determine its type, if volt/div = 2.5V. (Ans: unknown current = 5A,  AC current)

1. Calculate the unknown current in a circuit, if it is passed through a known value of resistor of 5Ω. The voltage drop obtained across the known resistor, is connected to y-input of CRO and a straight line above x-axis is observed having vertical displacement of 4.5cm with volt/div = 500mV. (Ans: current = +0.45A)
2. Calculate the r.m.s. value of current obtained in problem #1, above. (Ans: 3.54A)

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