Current Electricity

Current Electricity

Easy Questions

Electricity and Magnetism

Current Electricity
Moving Charges and Magnetism
Magnetism and Matter
Electromagnetic Induction
Alternating Current
Electromagnetic Waves
  1. In producing chlorine through electrolysis, 100 watt power at 125 V is being consumed. How much chlorine per minute is liberated? E.C.E. of chlorine is  0.367 × 10–6 kg/ coulomb.​​








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    Power = V × I

    I =
    Power
    		 =
    100
    V125

    E.C.E. of chlorine is 0.367 × 10–6 kg/coulomb
    Charge passing in one minute =
    100
    		 × 60 = 48 coulomb
    125

    Chlorine precipitated = 0.367 × 10–6 × 48
    17.6 × 10–6 kg
    = 17.6 mg

    Correct Option: D

    Power = V × I

    I =
    Power
    		 =
    100
    V125

    E.C.E. of chlorine is 0.367 × 10–6 kg/coulomb
    Charge passing in one minute =
    100
    		 × 60 = 48 coulomb
    125

    Chlorine precipitated = 0.367 × 10–6 × 48
    17.6 × 10–6 kg
    = 17.6 mg

  1. A steady current of 1.5 amp flows through a copper voltameter for 10 minutes. If the electrochemical equivalent of copper is 30 × 10–5 g coulomb–1, the mass of copper deposited on the electrode will be​








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    We have,  m = ZIt
    ​where, Z is the electrochemical equivalent of copper. ​
    ⇒ m = 30 x 10-5 x 1.5 x 10 x 60
    ​= 0.27 gm.

    Correct Option: C

    We have,  m = ZIt
    ​where, Z is the electrochemical equivalent of copper. ​
    ⇒ m = 30 x 10-5 x 1.5 x 10 x 60
    ​= 0.27 gm.

  1. A cell can be balanced against 110 cm and 100 cm of potentiometer wire, respectively with and without being short circuited through a resistance of 10Ω. Its internal resistance is








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    Here

    E >
    ER
    R + r

    hence the lengths 110 cm and 100 cm are interchanged.
    Without being short-circuited through R, only the battery E is balanced.
    E =
    V
    		 × l1 =
    V
    		 × 110 ................(i)
    LL

    When R is connected across E, Ri =
    V
    		 × l2
    L

    Or, R
    E
    		 =
    V
    		 × 100..............(ii)
    R + rL

    Dividing (i) by (ii), we get
    R + r
    		 =
    110
    R100

    or, 100 R + 100 r = 110 R

    Correct Option: A

    Here

    E >
    ER
    R + r

    hence the lengths 110 cm and 100 cm are interchanged.
    Without being short-circuited through R, only the battery E is balanced.
    E =
    V
    		 × l1 =
    V
    		 × 110 ................(i)
    LL

    When R is connected across E, Ri =
    V
    		 × l2
    L

    Or, R
    E
    		 =
    V
    		 × 100..............(ii)
    R + rL

    Dividing (i) by (ii), we get
    R + r
    		 =
    110
    R100

    or, 100 R + 100 r = 110 R

  1. A student measures the terminal potential difference (V) of a cell (of emf E and internal resistance r) as a function of the current (I) flowing through it. The slope and intercept, of the graph between V and I, then, respectively, equal:​​








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    The terminal potential difference of a cell is given by V + Ir = E ​ 
    V = VA – VB ​
    or  V = E – Ir
    ⇒ dV/dI = -r,
    Also for,  i = 0 then V = E ​
    ∴​slope = – r, intercept = E

    Correct Option: A

    The terminal potential difference of a cell is given by V + Ir = E ​ 
    V = VA – VB ​
    or  V = E – Ir
    ⇒ dV/dI = -r,
    Also for,  i = 0 then V = E ​
    ∴​slope = – r, intercept = E

  1. See the electric circuit shown in the figure.

    Which of the following equations is a correct equation for it?​​








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    Applying Kirchhoff ’s rule in loop abcfa
    ​ε1 – (i1 + i2) R – i1  r1 = 0.

    Correct Option: D


    Applying Kirchhoff ’s rule in loop abcfa
    ​ε1 – (i1 + i2) R – i1  r1 = 0.