Question

Calculate the wattage of soldering iron with a hot resistance $4K\Omega $ and operating voltage of$200V$.

Answer 1

Hint: Wattage here means power as power is measured in watts. The relation of power with resistance and voltage can be used to solve it.
(1) $P = \dfrac{{{V^2}}}{R}$
Where P is power
V is applied voltage or potential difference and R is resistance.

Complete step by step answer:
The power rating of an electrical appliances is the electrical energy consumed per second by the appliance when connected across the marked voltage mains
i.e. Power $ = \dfrac{W}{t}$…. (i)
When W is electrical energy consumed
t is the time take as potential difference, $V = \dfrac{W}{q}$
Where q is change.
So, $W = Vq$ … (ii)
Put (ii) in (i), we get
$P = \dfrac{{Vq}}{t} = VI$ … (iii) … (Where I is current and $I = q/t$)
Also, from ohm's law,
$
  V = IR \\
   \Rightarrow I = \dfrac{V}{R} \\
 $
Put this is equation (iii), we get
Power, $P = VI$
$
  P = V \times \dfrac{V}{R} \\
  P = \dfrac{{{V^2}}}{R} \\
 $
Now, here in the question,
Wattage, $P = ?$
Operating voltage, $V = 200V$
And resistance, $R = 4K\Omega = 4000\Omega $
So, power becomes
$
  P = \dfrac{{{V^2}}}{R} \Rightarrow P = \dfrac{{{{\left( {200} \right)}^2}}}{{4000}} \\
   \Rightarrow P = \dfrac{{40,000}}{{4000}} \Rightarrow P = 10watt. \\
 $
So, the wattage of soldering iron with a hot resistance $4K\Omega $ and operating voltage of $200V$ is $10$ Watt.

Additional Information:
The different formulas that can be used to find the power are,
(1) $P = VI$
(2) $P = \dfrac{{{V^2}}}{R}$ …. $\left( {as{\text{ V}} = IR \Rightarrow I = \dfrac{V}{R}{\text{ so, P}} = VI = V \times \dfrac{V}{R} = \dfrac{{{V^2}}}{R}} \right)$
(3) $P = {I^2}R$… (as $V = IR$ and $P = VI = IR \times R{\text{ P}} = {I^2}R$)
This can be used directly also to find the power.

Note:
It can also be solved by firstly finding the current from ohm’s law and then putting in $P = VI$ i.e.
From ohm’s law,
$V = IR$
As $V = 200v$ and $R = 4k\Omega = 4000\Omega $
So, $200V = I \times 4000\Omega $
$ \Rightarrow I = \dfrac{{200}}{{4000}}$ Ampere
So, current $I = \dfrac{1}{{20}}$ Ampere.
As, power, $P = VI$
$ = 200 \times \dfrac{1}{{20}}$
$P = 10$ Watt.