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

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Solution

(i) Given : $I = 400 m A = \frac{400}{1000} = 0.4 A, V = 12 V$
Resistance of the music system $= \frac{V}{1} = \frac{12}{0.4} = \frac{120}{4} = 30 Ω$ .
(ii) Given : $I = 320 m A = \frac{320}{1000} = 0.32 A$ , Resistance of music system $= 30 Ω$
According to Ohm's law,
$V = I R = 30 \times 0.32 = 9.60 V$
The music system stops playing at $9.60 V$ .

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(i) Given : I=400mA=4001000=0.4A,V=12VResistance of the music system =V1=120.4=1204=30Ω.(ii) Given : I=320mA=3201000=0.32A, Resistance of music system =30ΩAccording to Ohm's law,V=IR=30×0.32=9.60VThe music system stops playing at 9.60V.