Question

An electric device draws 6.50 A at 240 V. part aIf the voltage drops by 16 %, what will be the current, assuming nothing else changes?Express your answer using two significant figures.part bIf the resistance of the device were reduced by 16 % , what current would be drawn at 240 V?

Answer 1

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%

Step-by-step explanation:

According to Ohm's law,

`V\propto\text{ I}`

(a) So, if the voltage reduces by 16%, the current will also reduce by 16%.

The value of the current will be

`\begin{gathered} I_a=I-(I*(16)/(100)) \\ =6.5-(6.5*(16)/(100)) \\ =5.46\text{ A} \end{gathered}`

(b) The value of resistance initially is

`\begin{gathered} V=IR \\ R=(V)/(I) \\ =(240)/(6.5) \\ =36.92\text{ }\Omega \end{gathered}`

If the resistance is reduced by 16%, then the current can be calculated as

`\begin{gathered} I_b=(V)/(R^(\prime)) \\ =(V)/(R-(R*(16)/(100))) \\ =(240)/(36.92-(36.92*(16)/(100))) \\ =7.74\text{ A} \end{gathered}`

Thus, the current value decreases to 5.46 A when voltage decreases but the current value increases to 7.74 A when resistance decreases.