A music system draws a current of 400 mA 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 mA. At what battery voltage does the music system stop playing?

Given,

I = 400 mA = 400 x 10^-3 = 0.4 A

V = 12 V

From Ohm's law,

V = IR

Substituting the values in the formula we get,

$12 = 0.4 \times R \\[0.5em] \Rightarrow R = \dfrac{12}{0.4} \\[0.5em] \Rightarrow R = 30 Ω$

Hence, the resistance of the music system = 30 Ω

(b) Given,

I = 320 mA = 320 x 10^-3 A = 0.32 A

R = 30 Ω

From Ohm's law,

V = IR

Substituting the values in the formula we get,

V = 0.32 x 30 = 9.6V

Hence, the battery voltage when the music system stops playing = 9.6 V