Question

An electric geyser consumes \[2.2\]units of electrical energy per hour of its use. It is designed to work on the mains voltage of\[220V\].
a) What is the ‘power-rating’ of this device?
b) What is the current flowing through this device when it is connected across the ‘mains’?
c) What is the ‘resistance’ of this device?
d) Does the resistance of this device remain constant during its operation/working?
e) Find the cost of energy consumed if each unit cost\[Rs6\].

Answer 1

Hint: KWh is the commercial unit of energy which signifies the energy consumed by any electrical appliances in terms of Joule. This KWh is equivalent to units we consumed in our homes, offices, schools etc. The electricity bills are composed in terms of total units we consumed in a month.

Complete answer:
As we know that KWh is equivalent to units consumed by any electrical appliances.
So, it can be written in equation form as-
\[1unit=1Kwh\]\[=3.6\times {{10}^{6}}J\].
So,\[1unit\] consumes\[3.6\times {{10}^{6}}J\]of energy.
Here it is given in the question that geysers consume \[2.2\]units of electric energy.
So, \[2.2units=2.2Kwh=2.2\times 3.6\times {{10}^{6}}J\]\[=7.92\times {{10}^{6}}J\].
So, geysers consume \[7.92\times {{10}^{6}}J\]of energy in 1 hour.
a).Work = Energy Consumed by geyser = \[7.92\times {{10}^{6}}J\]
Time of operating the device \[=1hr=3600s\]
As we know that,
\[\because Power=\dfrac{Work}{Time}=\dfrac{Energy}{Time}\]
\[\therefore Power=\dfrac{7.92\times {{10}^{6}}}{3600}\]
\[Power=2200Watt\]
So the power rating of this device is\[2200Watt\].
SI unit of power is Watt.
b).Let us assume the value of current is\[I\].
Voltage of mains supply is given \[=220V\]
Power consumed by device \[=2200Watt\]
\[\because P=VI\]
\[I=\dfrac{P}{V}=\dfrac{2200}{220}=10A\]
So current flowing through this device when it is connected to mains is\[10A\].
SI unit of Current in Ampere \[(A)\]
c).Let us assume the resistance of this device is\[R\].
Operating Voltage of device \[V=220Volt\]
Current Flowing through device\[I=10A\]
\[\because R=\dfrac{V}{I}=\dfrac{220}{10}=22\Omega \]
So resistance of the device is\[22\Omega \].
SI unit of resistance is Ohm\[(\Omega )\]
d).As we know that Resistance and Temperature are directly proportional to each other, so if temperature of device will increase then its resistance will also increase and if it is decreased then resistance also gets decreased. So here the temperature of geysers will increase so its resistance to geysers will also increase. It can be expressed mathematically by the given equation.
\[{{R}_{t}}={{R}_{0}}(1+\alpha \Delta T)\]
Where,
\[{{R}_{t}}\]= Resistance of Conductor at\[{{t}^{0}}C\].
\[{{R}_{0}}\]= Resistance of Conductor at\[{{0}^{0}}C\]
\[\alpha \]= Temperature Coefficient.
\[\Delta T\]= Temperature difference.
e).As it is given in the question that ,
Cost of \[1unit=Rs6\]
 It is given that geysers consume \[2.2units\]of electrical energy.
So, Cost of \[2.2units=2.2\times 6\]\[=Rs13.20\]

Note:
The main point here is that when any electrical device is operating at mains supply it heats up and we know that resistance and temperature are directly proportional to temperature of that conducting device will always increase. In semiconductors, resistance and temperature is inversely proportional to each other so on increasing the temperature of a semiconductor its resistance will decrease.