Question

A series circuit consisting of an inductance free resistance $R=0.16\text{k}\Omega$ and a coil with active resistance is connected to the mains with effective voltage $V=220\text{V}$. The effective voltage across the resistance $R$ and the coil are equal to $V_1=80\text{V}\text{and}{V}_2=180\text{V}$ respectively. The value of $\frac{H}{10}\text{watt}$ where $H$ is the heat generated in the coil is

Answer 1

Effective mains voltage $V=220\text{V}$

$V_1=80\text{V};V_2=180\text{V}$

Current in the circuit

$I=\frac{V_1}{R}=\frac{80}{160}=0.5\text{A}$

Also, $I=\frac{V}{\sqrt{(R+r{)}^2+X_L^2}}$

Now, $V_1=IR=\frac{VR}{\sqrt{(R+r{)}^2+X_L^2}}$

$\Rightarrow (R+r{)}^2+X_L^2={\left(\frac{VR}{V_1}\right)}^2.........(1)$

And, $V_2=I\times \sqrt{r^2+X_L^2}$

$\Rightarrow V_2=\frac{V}{\sqrt{(R+r{)}^2+X_L^2}}\times \sqrt{r^2+X_L^2}$

From equation $\left(1\right)$, we get

$\Rightarrow V_2=\frac{V}{\frac{VR}{V_1}}\times \sqrt{r^2+X_L^2}$

$\Rightarrow V_2=\frac{V_1}{R}\times \sqrt{r^2+X_L^2}$

$\Rightarrow r^2+X_L^2={\left(\frac{V_{2}R}{V_1}\right)}^2..........\left(2\right)$

Subtracting $\left(2\right)$ from $\left(1\right)$

$\Rightarrow R^2+2Rr=\frac{R^2}{V_1^2}(V^2-V_2^2)$

$\Rightarrow r=\frac{R}{2V_1^2}(V^2-V_1^2-V_2^2)$

$\Rightarrow r=\frac{160}{2\times {80}^2}({220}^2-{80}^2-{180}^2)$

$\Rightarrow r=120\Omega$

Heat generated in the coil

$H=I^{2}r={0.5}^2\times 120=30\text{W}$

Hence, $\frac{H}{10}=3$