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Kirchhoff ’s laws and Potentiometer Test - 1

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Kirchhoff ’s laws and Potentiometer Test - 1
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  • Question 1
    1 / -0.25

    Consider the following two statements

    1. Kirchhoffs junction law follows from conservation of charge.

    2. Kirchhoffs loop law follows from conservative nature of electric field.

    Solution
    CONCEPT:

    There are two types of Kirchoff’s Laws:

    Kirchoff’s first law:

    • This law is also known as junction rule or current law (KCL). According to it the algebraic sum of currents meeting at a junction is zero i.e. Σ i = 0.

    • In a circuit, at any junction, the sum of the currents entering the junction must be equal to the sum of the currents leaving the junction i.e., i1 + i3 = i2 + i4
    • This law is simply a statement of “conservation of charge” as if the current reaching a junction is not equal to the current leaving the junction, the charge will not be conserved.

     

    Kirchoff’s second law:

    • This law is also known as loop rule or voltage law (KVL) and according to it “the algebraic sum of the changes in potential in complete traversal of a mesh (closed-loop) is zero”, i.e. Σ V = 0.
    • This law represents “conservation of energy” as if the sum of potential changes around a closed loop is not zero, unlimited energy could be gained by repeatedly carrying a charge around a loop.
    • Kirchhoff's loop law also follows the conservative nature of the electric field.
    • If there are n meshes in a circuit, the number of independent equations in accordance with the loop rule will be (n - 1).

     

    EXPLANATION:

    • Kirchhoff's junction law states that the current entering the junction is equal to the current leaving the junction, which indirectly means the charge entering the junction is equal charge leaving the junction. From the statement, it is clear that Kirchhoff's junction law follows from the conservation of charge. Therefore statement 1 is correct.
    • Kirchhoff's loop law states that the net potential drop in a loop is Zero Σ V = 0, which means potential changes around a closed loop is zero.
    • Voltage is the measure of energy per unit charge in this context KVL can be restated as the total energy gained per unit charge must be equal to the amount of energy lost per unit charge (ie: Law of conservation of Energy). Therefore statement 2 is correct.
  • Question 2
    1 / -0.25

     The current in the circuits is

    Solution

    CONCEPT:

    There are two types of Kirchoff’s Laws:

    Kirchoff’s first law:

    • This law is also known as junction rule or current law (KCL). According to it the algebraic sum of currents meeting at a junction is zero 
    •  Σ in = 0.

     

    • In a circuit, at any junction, the sum of the currents entering the junction must be equal the sum of the currents leaving the junction i.e.,
    •  i1 + i3 = i2 + i4
    • This law is simply a statement of “conservation of charge” as if the current reaching a junction is not equal to the current leaving the junction, the charge will not be conserved.

     

    Kirchoff’s second law:

    • This law is also known as loop rule or voltage law (KVL) and according to it “the algebraic sum of the changes in potential in the complete traversal of a mesh (closed-loop) is zero”,
    •  Σ Vn = 0.
    • This law represents “conservation of energy” as if the sum of potential changes around a closed loop is not zero, unlimited energy could be gained by repeatedly carrying a charge around a loop.
    • If there are n meshes in a circuit, the number of independent equations in accordance with the loop rule will be (n - 1).

     

    Explanation:

    Applying KVL in the given circuit:

    - 36 + 4io + 5io = 0

    9io = 36

    Io = 36/9

    io = 4 A

  • Question 3
    1 / -0.25
    Which of the following is not true about the wheat stone bridge?
    Solution

    CONCEPT:

    • Wheatstone bridge: It is an arrangement of four resistance which can be used to measure one of them in terms of rest. Here arms AB and BC are called ratio arm and arms AC and BD are called conjugate arms.
    • The bridge is said to be balanced when deflection in the galvanometer is zero i.e. no current flows through the galvanometer or in other words VB = VD.
    • In the balanced condition


    \(\frac{P}{Q} = \frac{S}{R}\)

    EXPLANATION:

    • The wheat stone bridge is used to measure the unknown resistance. Therefore, option 1 is correct.
    • Wheat stone bridge works on the principle of null deflection i.e. the ratio of their resistances is equal, and no current flows through the galvanometer. Thus option 2 is correct.
    • Meter bridge works on the principle of the wheat stone bridge. Thus option 3 is correct.
    • It is not used to measure the internal resistance of the battery. Thus option 4 is incorrect.
  • Question 4
    1 / -0.25
    Which value cannot be measured by potentiometer?
    Solution

    CONCEPT:

    • potentiometer is a device mainly used to measure the emf of a given cell and to compare emf of cells.
    • It is also used to measure the internal resistance of a given cell.
    • If Potential across AJ (V) is equal to emf E’ then no current will flow in the galvanometer circuit this condition to known as null deflection positionlength AJ (x) is known as balancing length.

    Potential gradient (k) :

    • The potential difference (or fall in potential) per unit length of wire is called potential gradient i.e. 

    \(k = \frac{V}{l}\)

    Where V = Potential difference and l = length of the wire

    EXPLANATION:

    • As we know K is the potential gradient of potentiometer wire, then the potential difference will be

    \(\Rightarrow V = KL \Rightarrow K = \frac{V}{L} = \frac{{IR}}{L} = I\frac{R}{L} = \left( {\frac{e}{{R + {R_h+r}}}} \right)\frac{R}{L}\)

    Where e = emf of the battery, Rh = resistance of the resistor, R = resistance of the wire, r = internal resistance of the battery, and L = length of the wire

    • From the above, it is clear that the potentiometer is used to measure the potential difference and the internal resistance of the battery. Therefore options 1 and 4 are correct.
    • The electric current can also be measured by using a potentiometer. Therefore option 3 is correct.
    • The capacity of the capacitance cannot be measured by using a potentiometer. Therefore option 2 is incorrect.
  • Question 5
    1 / -0.25

    As per the figure, which is the correct relation?

    Solution

    CONCEPT:

    Kirchhoff’s Law: Kirchhoff's circuit laws are two equalities that deal with the current and potential difference. They were first described in 1845 by German physicist Gustav Kirchhoff.

    • Kirchhoff’s Current Law (KCL): The algebraic sum of currents at any node of a circuit is zero. It can be described as also, the total current enters is exactly equal to the total current leaves the point.
      • It is also known as Kirchhoff’s first law.

    i1 + i2 + i3 + i4 + i5 = 0

    EXPLANATION:

    According to Kirchoff’s current law, the algebraic sum of current meeting at any junction in a circuit is zero. The current flowing towards the junction is positive and the current flowing away from the junction is negative.

    From the fig.

    I1 + I+ I6 = I+ I+ I2

    So option 2 is correct.

  • Question 6
    1 / -0.25

    Find the magnitude of resistance R in the circuit shown below, when no current flows through 5 Ω resistor.

    Solution

    CONCEPT:

    • Wheatstone bridge: It is an arrangement of four resistance which can be used to measure one of them in terms of rest. Here arms AB and BC are called ratio arm and arms AC and BD are called conjugate arms.
    • The bridge is said to be balanced when deflection in the galvanometer is zero i.e. no current flows through the galvanometer or in other words VB = VD.
    • In the balanced condition

     

    \(\frac{P}{Q} = \frac{S}{R}\)

    EXPLANATION:

    Given –

    As no current flows through 5 Ω resistors, the circuits represent a balanced Wheatstone bridge.

    \(\therefore \frac{{10}}{R} = \frac{{20}}{{10}} = 2\)

    \(\Rightarrow R = \frac{{10}}{2} = 5\;{\rm{\Omega }}\)

    ∴ Resistance R in the circuit is 5 Ω.

  • Question 7
    1 / -0.25
    Which instrument is placed at the center of a Wheatstone bridge?
    Solution

    CONCEPT:

    • Wheatstone bridge: It is an arrangement of four resistance which can be used to measure one of them in terms of rest. Here arms AB and BC are called ratio arm and arms AC and BD are called conjugate arms.
    • The bridge is said to be balanced when deflection in the galvanometer is zero i.e. no current flows through the galvanometer or in other words VB = VD.
    • In the balanced condition

    \(\frac{P}{Q} = \frac{S}{R}\)

    EXPLANATION:

    • The galvanometer is placed at the center of the Wheatstone bridge. Therefore option 2 is correct.
    • A Wheatstone bridge is a special arrangement of 4 resistors. It can be used to find an unknown resistance.
    • Also, if the Wheatstone bridge is balanced, there will be no current flowing through the galvanometer.
  • Question 8
    1 / -0.25

    Determine the potential at point B considering the potential at point A to be 0 in the following circuit.

    Solution

    CONCEPT:

    Kirchhoff’s Voltage Law (KVL): The algebraic sum of voltage (or voltage drops) in any closed path of a network in a single direction is zero.

    It is also known as Kirchhoff’s second law.

    V – V2 – V3 + V4 = 0

    V + V4 = V2 + V3

    Calculation:

    According to Kirchhoff’s  law along the path ACMB

    Va + 1V + 1 × 2 - 2 = Vb

    Given, V= 0,

    Hence Vb = 0 + 1 + 2 - 2 = 1V

    Important Points

    Kirchhoff’s Current Law (KCL): The algebraic sum of currents at any node of a circuit is zero. It can be described as also, the total current enters is exactly equal to the total current that leaves the point.

    It is also known as Kirchhoff’s first law.

     

    i1 + i2 + i3 + i4 + i5 = 0

  • Question 9
    1 / -0.25
    On what principle does Meter bridge work?
    Solution

    CONCEPT:

    • Meter bridge: Meter bridge is instruments based on the principle of Wheatstone bridge and is used to measure an unknown resistance.
    • In the case of a meter bridge, the resistance wire AC is 100 cm long. Varying the position of tapping point B, the bridge is balanced.
    • If in the balanced position of bridge AB = l, BC (100 – l) so that

     

    \(\frac{Q}{P} = \frac{{\left( {100 - l} \right)}}{l}\)

    Also, \(\frac{P}{Q} = \frac{R}{S} \Rightarrow S = \frac{{\left( {100 - l} \right)}}{l}R\)

    EXPLANATION:

    • A potentiometer is a device mainly used to measure emf of a given cell and to compare emf’s of cells. Therefore option 1 is incorrect.
    • The Meter Bridge is instruments based on the principle of Wheatstone bridge and is used to measure an unknown resistance. Therefore option 2 is correct.
    • Kirchoff's Laws are used to find the current and emf in the complicated circuits. Therefore option 3 is incorrect.
    • Ohm’s law gives the relation between the voltage and current. Therefore option 4 is incorrect.
  • Question 10
    1 / -0.25
    Which is the best device to measure e.m.f. of a cell:
    Solution

    CONCEPT:

    Some measuring devices are:

    S.NoVoltmeterAmmeterGalvanometerPotentiometer
    1.

    It is a device used to measure potential difference and is always put in parallel with the ‘circuit element’ across which potential difference is to be measured.

    It is the device to measure current and is always put in connected in series across the device the current is to be measured.

    An instrument used to detect, measure, and determine the direction of electric currents is called a galvanometer.

    potentiometer is a device mainly used to measure the emf of a given cell and to compare emf's of cells.

    2.

    Ideally, the voltmeter has infinite resistance.

    Ideally, the ammeter has zero resistance.A resistance of high value is connected in series to convert the galvanometer into a voltmeter.

    It is also used to measure the internal resistance of a given cell.

    EXPLANATION:

    • From the above, it is clear that a potentiometer is the best device to measure e.m.f. of a cell. Therefore option 4 is correct.
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