Question

A music system draws a current of $400mA$ when connected to a $12V$ battery.
(i) What will be the resistance of the music system?
(ii) The music system is left playing for so many hours and finally the battery voltage gets dropped and the music system stops playing when the current drops to $320mA$. At which voltage of the battery does the music system stop playing?

Answer 1

Hint: The basic concept behind the solution of the question is Ohm’s law. As the voltage of a circuit increases, the current flowing through the wire will also get increased. The proportionality constant is known as the resistance. The resistance will be a constant for a definite length and area of cross section. First of all, find the resistance of the music system. Then find out the voltage at which the system stops playing music using ohm’s law.

Complete answer:
Let us take a look at the values given in the question.
The current drew by the music system is given as,
$I=400mA$
Voltage of the battery connected is given as,
$V=12V$
Therefore the resistance of the music system will be given as,
$R=\dfrac{V}{I}$
Substituting the values in it will give,
$R=\dfrac{12V}{400mA}=30\Omega $
According to the ohm’s law,
$V=IR$
The voltage of the system will be given as the product of current passing through the conductor and resistance of the wire.
 Substituting the values of current and resistance in it,
$V=30\times 0.32=9.60V$
Therefore the music system stops playing at $9.6V$.
Therefore the answer is obtained.

Note:
Resistance of the circuit is defined as the opposition provided by a wire for the flow of electric current through the system. It will be proportional to the length of the wire. The resistance is also found to be inversely proportional to the area of cross section.