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    3 problems on Electrinc Current: Ohm's law

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    SOLUTION This solution is FREE courtesy of BrainMass!

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    Answer:

    In the series circuit the current I through all the resistors is same. This means that any of the resistor is not open and the current through each will be same as through R2. Ohm's law as gives the voltage across the resistor as

    V = R*I

    As voltage across R2 is non-zero, a current is there through R2. This implies that there must be the same current through the whole circuit of through every resistance, and hence the voltage across all resistances must be non-zero.

    But we see that there is no voltage shown across four resistors, this may happen because any of the reasons

    1. The resistors are short-circuited.
    2. The resistors having zero resistance (good conductor)
    3. The voltmeter is not connected properly at these resistors.

    The most appropriate reason is that the resistances are short-circuited and hence these resistors do not oppose the current. As the current is only opposed by R2, the whole voltage drop is across it and hence the voltage measured across R2 will be 12 Volts.

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    Answer to Q 51:

    a) 1) The voltage shown by the voltmeter R4 is 12 V means there is current in the circuit and as the voltage drop across R4 is 12 V, same as emf of the battery, all other resistors are short-circuited, moreover as no current is measured by ammeter, it is faulty.

    2) There may be other reason that the voltmeter may be faulty and jammed to 12 V mark and any of the components is open so that no current is flowing through the circuit.
    b) The ammeter is showing the current but the voltmeter is showing no voltage drop means that either the voltmeter is faulty or the resistances R4 and R5 are short-circuited.

    If the voltmeter is faulty then the ammeter reading must be
    I = V/R = 10/500 = 20 mA.
    But the current shown by ammeter is 33.3 mA.

    If the resistors R4 and R5 are short-circuited the current in the circuit will be
    I = V/R = 10/300 = 33.3 mA
    Which is the current shown by ammeter and hence we conclude that R4 and R5 are short-circuited.

    Answer to Q 52:

    If only R2 is shorted and R4 and R5 are already short-circuited, the equivalent resistance of the circuit will be that of only R1 and R3, connected in series, which is 200 Ω. Hence the current in the circuit will be

    I' = V/R = 10/200 = 0.05 A = 50 mA.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 5:47 pm ad1c9bdddf>
    https://brainmass.com/physics/ohms-law/problems-electrinc-current-ohms-law-70759

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