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Four 15 Ω resistors are connected in series with a 45 V battery. What is the current in the circuit?

Sagot :

Answer:

3A

Explanation:

Resistance=R=15

Voltage=45volts

  • we know that

[tex]\boxed{\sf Current=\dfrac{Voltage}{Resistance}}[/tex]

[tex]\\ \tt \mapsto \:Current=\dfrac{45}{15}[/tex]

[tex]\\ \tt \mapsto \:Current=\dfrac{3A}[/tex]

Ritmeksidn

The current of the circuit is 0.75 amps

The first step is to write out the parameters

Four resistors of 15 in series

15+15+15+15

= 60

voltage= 45

Current= 45/60

= 0.75

Hence the current is 0.75 amps

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