# Measuring Resistance

Topics: Power, Energy, Electric current Pages: 4 (649 words) Published: February 20, 2013
Measuring Resistance Experiment Evaluation

M U S (12037486)

Title Page1
Introduction3
Research Questions4
Questions4
Experiment6

Introduction

This experiment is carried out to show how the

Research Questions

Questions

1) Provide the units used to measure the following quantities, electric current, inductance, frequency, power and energy (work).

2) What is the difference between AC and DC voltages? Give examples.

3) Provided a household has:
* 10 light bulbs of 60 watts each (used 6h/day on average) * TV using 100 watts (used 4h/day on average)
* A washing machine using 1000 watts (used for 3 /h once every 3 days) Calculate the total average energy consumption of this household in a quarter (90 days) in joules than in Kilowatt hours (kWh).

4) If the energy consumption is charged at 20p/kWh, what is the cost of the electricity bill for that quarter?

1) The following quantities are measured in the following units; - Electric Current is measured in Ampere, A.
- Inductance is measured in Henry, H.
- Frequency is measured in Hertz, Hz.
- Power is measured in Watts, W.
- Energy (work) is measured in Joules, J.

2) AC

3) Power = Energy / Time, therefore Energy = Power x Time. i) 10 bulbs x 60 Watts = 600 Watts, 600 Watts x 6 Hours = 3600 Wh = 3.6 kWh per day. ii) 100 Watts x 4 Hours = 400 Watts = 0.4 kWh per day.

iii) 1000 Watts x 3 Hours = 3000 Wh every 3 days = 1 kWh per day. iv) 1 kWh + 0.4 kWh + 3.6 kWh = 5 kWh per day.
v) 5 kWh x 90 Days = 450 kWh per quarter (90 days).
vi) Using 1 Joule/sec = 1 watt, 1 Joule = 1 Watt x sec, 1k/w = 1000 Watts & 3600 sec in 1 Hour; 450 kWh x 3600 x 1000 = 1620000000 = 1.62 x 10^9 Joules.

4) If electricity bill is 20p/kWh then 450 x 20p = £90.00 for the quarter.

Experiment

Experimental Equipment

* Ohmmeter

* 50 Identical 220 kΩ Resistors in theory...