To measure current using a digital multimeter the probes of the meter would be p

Question

To measure current using a digital multimeter the probes of the meter would be placed A) in parallel with B) in series with C) adjacent to U1sQ2 A reading of 2.80 is displayed when measuring the resistance of a resistor. The range setting is set to 20k. What is the resistance of the resistor? A) 2800 ? B) 2.80 2 C)5.60 ? D) 5600 2 U15Q3 A waveform displayed on an oscilloscope is measured to have 6.3 vertical divisions and 4.7 horizontal divisions for one complete cycle. The Volt/div knob is set to 0.2 V/div and the Time div knob to 0.5ms/div. Determine the peak to peak voltage of the waveform. A) 3.15 V B) 2.35 V C)0.94 V D) 1.26 V U15P1 Determine the period, peak to peak voltage, rms voltage and frequency for the waveform in the below picture. The Time div setting is 20us and the Volt/div setting is 1V/div period- peak to peak voltage- rms voltage frequency _ channel 1channel 2

Explanation / Answer

1. To measure current, Multimeter should become a part of the circuit, i.e it must be connected in series with the component.

2. 20k means...this setting can show maximum resistnresof 20KOhm.

On this setting, 2.80 shows, 2800Ohms.

3. Volts/divisions times the divisions in 1 cycle gives the volts.

I.e 6.3×0.2 = 1.26V peak to peak.

As voltage is on Vertical knob.

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