Question

$\left( i \right)$ What is the meaning of an electrical appliance having $2200W$ and $220V$ ?
$\left( {ii} \right)$ What is the resistance of this appliance?
$\left( {iii} \right)$ If the supply potential is $110V$ , What will be the power consumed by this appliance?
$\left( {iv} \right)$If this appliance is used for $8$ hours a day, what will be the electrical energy consumed?

Answer 1

Hint : As the power and voltage is given in the question, the resistance and electrical energy can be calculated by using the formula given below. In the third sub question the power should be calculated by using the resistance value obtained in the second sub question because the resistance of any appliance remains the same even if the power and voltage is changed.


Complete step-by-step solution:
Given:
Power, $P = $ $2200W$
Voltage, $V = $ $220V$
$\left( i \right)$ An electrical appliance having $2200W$ and $220V$ means the appliance will deliver a power of $2200W$ at voltage $220V$ . That is maximum power that is delivered by appliance will be $2200W$ at voltage $220V$ that can give a maximum current of
$I = \dfrac{P}{V}$
$I = \dfrac{{2200}}{{220}} = 10A$
That is the safe current the appliance can withstand is $10A$ .
$\left( {ii} \right)$ Resistance of the appliance:-
We know that,
 $R = \dfrac{{{V^2}}}{P}$
On substituting the given data in above equation
$R = \dfrac{{{{\left( {220} \right)}^2}}}{{2200}}$
On simplifying the above equation, we get
Therefore Resistance, $R = 22\Omega $
$\left( {iii} \right)$ The power consumed when voltage is $110V$
Given: $V = 110V$
Take, $R = 22\Omega $ because the resistance of appliance remains the same even the voltage is different
$P = \dfrac{{{V^2}}}{R}$
On substituting the given data in above equation
$P = \dfrac{{{{\left( {110} \right)}^2}}}{{22}}$
On simplifying the above equation, we get
Therefore power, $P = 550W$
$\left( {iv} \right)$
$ \Rightarrow $ Electrical energy in joules $\left( J \right)$
Given: $P = 2200$
$t = 8 \times 3600 = 28800s$
$E = P \times t$
On substituting the given data in above equation
$E = 2200 \times 28800$
On simplifying the above equation, we get
Therefore, Electrical energy in joules $E = 63360000J$
$ \Rightarrow $ Electrical energy in kilowatt hour $\left( {kwhr} \right)$
Given: $P = 2200$
$t = 8hr$
$E = \dfrac{{P \times t}}{{1000}}$
On substituting the given data in above equation
$E = \dfrac{{2200 \times 8}}{{1000}}$
On simplifying the above equation, we get
Therefore, Electrical energy in kilowatt hour $E = 17.6kwhr$

Note:It should be noted that if energy is required in joules $\left( J \right)$ the power should be substituted in watt and time should be substituted in seconds and if energy is required in kilowatt hour $\left( {kwhr} \right)$ the power should be substituted in watt and time should be substituted in hour in the equation $\left( 2 \right)$ and $\left( 3 \right)$ respectively.