Three resisters 5000 500 50 ohm are connected in series with the 555 voltage main what is the current

Question

Three resisters 5000 500 50 ohm are connected in series with the 555 voltage main what is the current

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Jade 4 weeks 2021-08-16T06:33:37+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-08-16T06:34:39+00:00

    Answer:

    R1= 5000ohms R2= 500ohms  R3= 50ohms

    Rresultant = R1+R2+R3

    R resultant = 5000+500+50

    R resultant = 5550ohms

    v = 555v

    I = V\R

    I= 555\ 5550

    I = 1\10 AMPERE

    Explanation:

    0
    2021-08-16T06:34:42+00:00

    Answer:

    0.1 A

    Explanation:

    As per the provided information in the given question, we have :

    • Three resistors of 5000Ω, 500Ω and 50Ω are connected in series.
    • Voltage or potential difference is 550 V.

    We are asked to calculate how much current is flowing in the circuit.

    In order to calculate how much current is flowing in the circuit, we need to find the equivalent resistance in the circuit.

    Since the resistors are connected in series combination, we need to apply the formula of equivalent resistance in series combination.

    As we know that, when resistors are connected in series combination, then equivalent resistance pf the circuit is given by,

     \twoheadrightarrow \quad \sf { R_S = R_1 + R_2 + R_3 + \dots R_n}

    Here,

    •  \rm R_1 = 5000Ω
    •  \rm R_2 = 500Ω
    •  \rm R_3 = 50Ω

    Substituting the values, we get:

     \twoheadrightarrow \quad \sf { R_S = R_1 + R_2 + R_3}

     \twoheadrightarrow \quad \sf { R_S = (5000 + 500 +50) \; \Omega}

     \twoheadrightarrow \quad \bf \underline { R_S = 5550 \; \Omega}

    Now, let’s calculate the current by using ohm’s law. We know that, according to Ohm’s law :

     \twoheadrightarrow \quad \sf { V = IR}

    • V denotes potential difference
    • I denotes current
    • R denotes resistance

    Substituting values,

     \twoheadrightarrow \quad \sf { 555 = I \times 5550}

     \twoheadrightarrow \quad \sf { \cancel{\dfrac{555}{5550}} \; A= I }

     \twoheadrightarrow \quad \sf {  \dfrac{1}{10} \; A= I }

     \twoheadrightarrow \quad \bf \underline{ 0.1 \; A= I }

    Therefore, 0.1 Ampere of current is flowing in the circuit.

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