Given:
The current is I = 6.5 A
The voltage is V = 240 V
To find the value of current
(a) if the voltage is reduced by 16%
(b) if resistance is reduced by 16%
Explanation:
According to Ohm's law,
[tex]V\propto\text{ I}[/tex](a) So, if the voltage reduces by 16%, the current will also reduce by 16%.
The value of the current will be
[tex]\begin{gathered} I_a=I-(I\times\frac{16}{100}) \\ =6.5-(6.5\times\frac{16}{100}) \\ =5.46\text{ A} \end{gathered}[/tex](b) The value of resistance initially is
[tex]\begin{gathered} V=IR \\ R=\frac{V}{I} \\ =\frac{240}{6.5} \\ =36.92\text{ }\Omega \end{gathered}[/tex]If the resistance is reduced by 16%, then the current can be calculated as
[tex]\begin{gathered} I_b=\frac{V}{R^{\prime}} \\ =\frac{V}{R-(R\times\frac{16}{100})} \\ =\frac{240}{36.92-(36.92\times\frac{16}{100})} \\ =7.74\text{ A} \end{gathered}[/tex]Thus, the current value decreases to 5.46 A when voltage decreases but the current value increases to 7.74 A when resistance decreases.
