Question

Calculate the wattage of soldering iron with a hot resistance $4K\mathrm{\Omega }$ 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) $P={\displaystyle \frac{V^2}{R}}$
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 $={\displaystyle \frac{W}{t}}$…. (i)
When W is electrical energy consumed
t is the time take as potential difference, $V={\displaystyle \frac{W}{q}}$
Where q is change.
So, $W=Vq$ … (ii)
Put (ii) in (i), we get
$P={\displaystyle \frac{Vq}{t}}=VI$ … (iii) … (Where I is current and $I=q/t$)
Also, from ohm's law,
$$\begin{gathered} V=IR \\ \Rightarrow I={\displaystyle \frac{V}{R}} \end{gathered}$$
Put this is equation (iii), we get
Power, $P=VI$
$$\begin{gathered} P=V\times {\displaystyle \frac{V}{R}} \\ P={\displaystyle \frac{V^2}{R}} \end{gathered}$$
Now, here in the question,
Wattage, $P=?$
Operating voltage, $V=200V$
And resistance, $R=4K\mathrm{\Omega }=4000\mathrm{\Omega }$
So, power becomes
$$\begin{gathered} P={\displaystyle \frac{V^2}{R}}\Rightarrow P={\displaystyle \frac{{\left(200\right)}^2}{4000}} \\ \Rightarrow P={\displaystyle \frac{40,000}{4000}}\Rightarrow P=10watt. \end{gathered}$$
So, the wattage of soldering iron with a hot resistance $4K\mathrm{\Omega }$ 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) $P={\displaystyle \frac{V^2}{R}}$ …. $\left(as\text{ V}=IR\Rightarrow I={\displaystyle \frac{V}{R}}\text{ so, P}=VI=V\times {\displaystyle \frac{V}{R}}={\displaystyle \frac{V^2}{R}}\right)$
(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 $R=4k\mathrm{\Omega }=4000\mathrm{\Omega }$
So, $200V=I\times 4000\mathrm{\Omega }$
$\Rightarrow I={\displaystyle \frac{200}{4000}}$ Ampere
So, current $I={\displaystyle \frac{1}{20}}$ Ampere.
As, power, $P=VI$
$=200\times {\displaystyle \frac{1}{20}}$
$P=10$ Watt.