Question

Two electric lamps are rated for $220volts60watts$, and $220volts40watts$. Find the heat generated in each lamp per second when they are connected in series across $220volts$.

Answer 1

Step 1: Given data

Two lamps in voltage $V=220volts$ having power

$$\begin{gathered} P_1=60watts \\ {P}_2=40watts \end{gathered}$$,

Let $H_1$ and $H_2$ be the heat generated by two lamps when connected in series across $220volts$.

Step 2: Formula used

Power of lamp having resistance $R$ across a potential $V$ is

$$\begin{gathered} \Rightarrow P=\frac{V^2}{R} \\ \Rightarrow R=\frac{V^2}{P} \end{gathered}$$

Step 3: Determining the equivalent resistance

For the two balls, the resistance will be,

$\begin{array}{rcl}{R}_1& =& \frac{V^2}{P_1}\\ & =& \frac{{\left(220\right)}^2}{60}\\ & =& 806.6\Omega \\ & & \end{array}$

and

$\begin{array}{rcl}{R}_2& =& \frac{V^2}{P_2}\\ & =& \frac{{\left(220\right)}^2}{40}\\ & =& 1210\Omega \end{array}$

As they are connected in series the equivalent resistance will be

$\begin{array}{rcl}R& =& R_1+R_2\\ & =& 806.6+1210\\ & =& 2016.6\Omega \end{array}$

Step 4: The current in the series is given by

$\begin{array}{rcl}I& =& \frac{V}{R}\\ & =& \frac{220}{2016.6}\\ & =& 0.1A\end{array}$

Now the heat produced($H$) in an element of resistance($R$) is given by,

$H=I^{2}RT$

where$T$ is the time

$I$ is the amount of current flowing through the circuit.

How heat generated in each bulb per time $T=1sec$ is,

$\begin{array}{rcl}{H}_1& =& I^2{R}_{1}t\\ & =& {(0.1)}^2\times (806.6)\times 1\\ & =& 8.06Joules\end{array}$

and

$\begin{array}{rcl}{H}_2& =& I^2{R}_{2}t\\ & =& {(0.1)}^2\times 1210\times 1\\ & =& 12.1Joules\end{array}$

Therefore, the heat generated in each lamp will be $8.06Joules$ and $12.1Joules$.