A 5.00 Ω, a 10.0 Ω, and a 15.0 Ω resistor are connected in series across a 90.0 V battery. What is the current in the cir

Question

A 5.00 Ω, a 10.0 Ω, and a 15.0 Ω resistor are connected in
series across a 90.0 V battery. What is the current in the
circuit?
a 5A
b
10 A.
с ЗА
d 8A

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Thu Giang 4 years 2021-07-25T14:39:45+00:00 1 Answers 13 views 0

Answers ( )

  1. vankhanh
    0
    2021-07-25T14:41:44+00:00
    Reply

    Answer: Option C) 3A

    Explanation:

    Given that,

    Voltage = 90.0 Volts

    Current I = ?

    R1 = 5.0Ω

    R2 = 10.0Ω

    R3 = 15.0Ω

    Since the resistors are connected in series, the total resistance (Rtotal) of the circuit is the sum of each resistance.

    i.e Rtotal = R1 + R2 + R3

    Rtotal = 5.0Ω + 10.0Ω + 15.0Ω = 30.0Ω

    Then, apply the formula for ohms law

    Voltage = Current x resistance

    V = I x Rtotal

    90.0V = I x 30.0Ω

    I = 90.0V / 30.0Ω

    I = 3 Amps (Amps is the unit of current)

    Thus, the current in the circuit is 3 Amps

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