Question

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?​

Answer 1

$\huge\purple{\boxed{\mathscr{✮Answer࿐}}}$

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(i)

☆Given:

→ $I=400 mA=\frac{400}{1000}=0.4A$

→ $V=12V$

☆Resistance of the music system:

⇒ $\frac{V}{I} =\frac{12}{0.4} =\frac{120}{4} =30Ω$

(ii)

☆Given:

→ $I=320mA=\frac{320}{1000} =0.32A$

→ Resistance of music system $= 30Ω$

☆The music system stops playing at :

⇒ $V=IR=30$×$0.32=9.60V$

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Hope it Helps !!!

Answer 2

Answer:

${\huge{\overbrace{\underbrace{\purple{Namaste }}}}}$

Explanation:

$\huge\pink{AN WER}$

Given:

→ I=400 mA=\frac{400}{1000}=0.4AI=400mA=

1000

400

=0.4A

→ V=12VV=12V

☆Resistance of the music system:

⇒ \frac{V}{I} =\frac{12}{0.4} =\frac{120}{4} =30Ω

I

V

=

0.4

12

=

4

120

=30Ω

(ii)

☆Given:

→ I=320mA=\frac{320}{1000} =0.32AI=320mA=

1000

320

=0.32A

→ Resistance of music system = 30Ω=30Ω

☆The music system stops playing at :

⇒ V=IR=30V=IR=30 ×0.32=9.60V0.32=9.60V

hope this will help you ❤️