Two resistors, R 1 = 2.01 Ω R1=2.01 Ω and R 2 = 5.29 Ω R2=5.29 Ω , are connected in series to a battery with an EMF of 24.0 24.0 V and negli

Question

Two resistors, R 1 = 2.01 Ω R1=2.01 Ω and R 2 = 5.29 Ω R2=5.29 Ω , are connected in series to a battery with an EMF of 24.0 24.0 V and negligible internal resistance. Find the current I 1 I1 through R 1 R1 and the potential difference V 2 V2 across R 2 R2 .

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bonexptip 4 years 2021-08-12T10:40:52+00:00 2 Answers 19 views 0

Answers ( )

  1. thanhha
    0
    2021-08-12T10:41:55+00:00
    Reply

    Answer:

    I1 = 3.288 A,

    V2 = 17.39 V

    Explanation:

    The combined resistance of the two resistor connected in series is given as

    R’ = R1+R2……………… Equation 1

    Where R’ = Combined resistance.

    Given: R1 = 2.01 Ω, R2 = 5.29 Ω

    Substitute into equation 1

    R’ = 2.01+5.29

    R’ = 7.3 Ω

    Using

    E = I(R’+r)……………. Equation 2

    Where E = emf of the battery, I = current through the circuit, r = internal resistance.

    Given: E = 24 V, R = 7.3 Ω, r = 0 Ω( Negligible)

    Substitute into equation 2

    24 = I(7.3)

    I = 24/7.3

    I = 3.288 A.

    Since the resistors are connected in series, the same amount of current flows through them

    Therefore,

    I = I1 = 3.288 A.

    Using ohm’s law,

    V2 = IR2……………… Equation 3

    Where V2 = potential difference across R2 resistor.

    Given: I = 3.288 A, R2 = 5.29 Ω

    Substitute into equation 3

    V2 = 3.288(5.29)

    V2 = 17.39 V

  2. ngocdiep
    0
    2021-08-12T10:42:02+00:00
    Reply

    Answer:

    I = 3.3 A

    V2 = 17.4 V

    Explanation:

    a)

    • In a circuit series, the current is the same through any point of the circuit.
    • Assuming the resistors are in the linear zone of operation, we can apply Ohm’s Law to both resistors, as follows:

            V = I* r_{eq} = V_{R1} + V_{R2} = (I*R_{1}) + (I*R_{2}) = I * (R_{1} + R_{2})\\ \\ R_{eq} = R_{1} + R_{2}

    • Therefore, we can find the current I as follows:

            I =\frac{V}{R_{eq} } = \frac{24.0 V}{7.3 \Omega} = 3.3 A

    b)

    • Applying Ohm’s law to R2, we can find the voltage through R2 as follows:

           V_{R2} = I* R_{2} = 3.3 A * 5.29 \Omega = 17.4 V

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