Question

Three identical resistances are connected in a star fashion against a balanced three phase voltage supply. If one of the resistances be removed then the percentage reduction in power is

• A. 17.32%
• B. 33.33%
• C. 14.14%
• D. 50%

Answer 1

The correct option is D 50%

When all the resistances are present;
$P=\frac{3V_{ph}^2}{R}=3{\left(\frac{V_L}{\sqrt{3}}\right)}^2\times \frac{1}{R}=\frac{V_L^2}{R}$
When one of the phase resistances is removed,
$P=2{\left(\frac{V_L^2}{2}\right)}^2\times \frac{1}{R}=\frac{V_L^2}{2R}$
$\therefore$ Reduction in power$=\frac{\frac{V_L^2}{R}-\frac{V_L^2}{2R}}{\frac{V_L^2}{R}}\times 100\%=50\%$