Current Electricity

More Topic from Physics

Question 1

A 5V battery with internal resistance 2Ω and a 2V battery with internal resistance 1Ω are connected to a 10Ω resistor as shown in the figure. The current in the 10 Ω resistor is 

  • 0.27 A P2 to P1

  • 0.03 A P1 to P2

  • 0.03 A P2 to P1

  • 0.27 A P1 to P2

Solution

C.

0.03 A P2 to P1


straight V subscript straight P subscript 2 end subscript minus straight V subscript straight P subscript 1 end subscript space equals space fraction numerator begin display style 5 over 2 end style plus begin display style 0 over 10 end style minus begin display style 2 over 1 end style over denominator begin display style 1 half end style plus begin display style 1 over 10 end style plus begin display style 1 over 1 end style end fraction
straight I space equals space straight V subscript straight P subscript 2 minus straight V subscript straight P subscript 1 end subscript end subscript over 10 space equals space 0.03 space from space straight P subscript 2 rightwards arrow straight P subscript 1
Question 2

A long straight wire of radius ‘a’ carries a steady current i. The current is uniformly distributed across its cross section. The ratio of the magnetic field at a/2 and 2 a is 

  • 1/4

  • 4

  • 1

  • 1/2

Solution

C.

1

straight B 2 straight pi straight a over 2 space equals space straight mu subscript 0 straight i over πa squared open parentheses πa squared over 4 close parentheses
straight B subscript 1 space equals fraction numerator straight mu subscript 0 straight i over denominator 4 πa end fraction space... left parenthesis straight i right parenthesis
straight B subscript 2 space 2 straight pi space left parenthesis 2 straight a right parenthesis space equals space straight mu subscript 0 space straight i
straight B subscript 2 space equals fraction numerator straight mu subscript 0 space straight i over denominator 4 space πa end fraction space space space... space space left parenthesis ii right parenthesis
straight B subscript 1 over straight B subscript 2 space equals 1
Question 3

A material ‘B’ has twice the specific resistance of ‘A’. A circular wire made of ‘B’ has twice the diameter of a wire made of ‘A’. Then for the two wires to have the same resistance, the ratio

  • 2

  • 1

  • 1/4

  • 1/2

Solution

A.

2

straight R subscript 1 space equals space fraction numerator straight rho subscript straight A calligraphic l subscript straight A over denominator πR subscript straight A superscript 2 end fraction
straight R subscript 2 space equals space fraction numerator straight rho subscript straight B calligraphic l subscript straight B over denominator πR subscript straight B superscript 2 end fraction
calligraphic l subscript straight A over calligraphic l subscript straight B space equals space fraction numerator straight rho subscript straight B straight R subscript straight A superscript 2 over denominator straight rho subscript straight A straight R subscript straight B superscript 2 end fraction space
equals space fraction numerator 2 straight rho subscript straight A space straight R subscript straight A superscript 2 over denominator straight rho subscript straight A 4 straight R subscript straight A superscript 2 end fraction
rightwards double arrow calligraphic l subscript straight A over calligraphic l subscript straight B space equals space 2 space space
Question 4

A material ‘B’ has twice the specific resistance of ‘A’. A circular wire made of ‘B’ has twice the diameter of a wire made of ‘A’. Then for the two wires to have the same resistance, the ratio AA /AB of their respective lengths must be

  • 2

  • 1

  • 1/4

  • 3

Solution

A.

2

straight R subscript 1 space equals space fraction numerator straight rho subscript straight A calligraphic l subscript straight A over denominator πR subscript straight A superscript 2 end fraction
straight R subscript 2 space equals fraction numerator straight rho subscript straight B calligraphic l subscript straight B over denominator πR subscript straight B superscript 2 end fraction
fraction numerator
calligraphic l subscript A over denominator calligraphic l subscript B end fraction space equals space space fraction numerator straight rho subscript straight B straight R subscript straight A superscript 2 over denominator straight rho subscript straight A straight R subscript straight B superscript 2 end fraction space
equals space fraction numerator 2 straight rho subscript straight B straight R subscript straight A superscript 2 over denominator straight rho subscript straight A 4 straight R subscript straight B superscript 2 end fraction space
fraction numerator
calligraphic l subscript A over denominator calligraphic l subscript B end fraction space equals 2
Question 5

A rectangular loop has a sliding connector PQ of length

  • straight I subscript 1 space equals space straight I subscript 2 space space equals space fraction numerator straight B space calligraphic l straight v over denominator 6 straight R end fraction comma space fraction numerator straight B space calligraphic l straight v over denominator 3 straight R end fraction
  • straight I subscript 1 space equals space minus straight I subscript 2 space equals space fraction numerator straight B space calligraphic l straight v over denominator straight R end fraction comma space straight I space equals fraction numerator 2 straight B space calligraphic l straight v over denominator 3 straight R end fraction
  • straight I subscript 1 space equals space straight I subscript 2 space space equals space fraction numerator straight B space calligraphic l straight v over denominator 3 straight R end fraction comma space straight I space equals space fraction numerator 2 straight B space calligraphic l straight v over denominator 3 straight R end fraction
  • straight I subscript 1 space equals straight I subscript 2 space equals space straight I space equals space fraction numerator straight B space calligraphic l straight v over denominator straight R end fraction

Solution

C.

straight I subscript 1 space equals space straight I subscript 2 space space equals space fraction numerator straight B space calligraphic l straight v over denominator 3 straight R end fraction comma space straight I space equals space fraction numerator 2 straight B space calligraphic l straight v over denominator 3 straight R end fraction

A moving conductor is equivalent to a battery of emf
= vBI
Equivalent circuit I = I1 + I2

Applying Kirchhoff's law
I1R + IR -vBl = 0 .... (i)
I2R + IR -vBl = 0 .... (ii)
Adding Eqs. (i) and (ii), we get
2IR + IR = 2vBI
I = 2VBI/3R
I1 = I2 = VBI/3R