Q: Which part of the curve shown in fig represents the postitive resistane?

Resistance=\dfrac{1}{slope} Region of positive slope represents positive resistance. In the given figure, AB,BC and DE represents positive resistance.

Q: In a potentiometer using two cells in series gives a balance length of 600cm . When the same cells are connected opposing each other then balance length is 100cm . The ratio of emf of larger to smaller cell is:

Case 1 \Rightarrow \left( { E }_{ 1 }+{ E }_{ 2 } \right) =I\times r\left( 600 \right) Case 2 \Rightarrow \left( { E }_{ 1 }-{ E }_{ 2 } \right) =I\times r\times100 \Rightarrow \dfrac { { E }_{ 1 }+{ E }_{ 2 } }{ { E }_{ 1 }-{ E }_{ 2 } } =\dfrac { 600 }{ 100 } { E }_{ 1 }+{ E }_{ 2 }={ 6E }_{ 1 }-{ 6E }_{ 2 } 7{ E }_{ 2 }=5{ E }_{ 1 } \Rightarrow \dfrac { { E }_{ 1 } }{ { E }_{ 2 } } =7/5

Question 1

Which part of the curve shown in fig represents the postitive resistane?

Accepted Answer

Resistance=\dfrac{1}{slope}  &#10;&#10;Region of positive slope represents positive&#10;resistance.&#10;&#10;In the given figure,&#10;AB,BC and DE represents positive resistance.

Question 2

The following table gives the current  i  through two devices for several values of potential difference V of Device 1 Device 2 (SI Units).Device 1Device 1Device 2Device 2 V I V I 24.521.536.7532.24942.8

Accepted Answer

Ohmic Resistance: A resistance can be said to ohmic resistance when the ratio of voltage to instantaneous current is constant.ieFor ohmic law     R=\dfrac {  V}{ I } = constant Non-Ohmic Resistance: A resistance can be said to Non-ohmic resistance when the ratio of voltage to instantaneous current is not constant.ieFor ohmic law     R
e\dfrac {  V}{ I } = constant Device 1                                Device 2 (1)\dfrac { 2 }{ 4.2 }  =\dfrac {  4}{ 9 }  \ \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \  (1)\dfrac { 2 }{ 1.5 } =\dfrac { 4 }{ 3 }  (2)\dfrac {  3}{ 675 }  =\dfrac {  4}{ 9 }   \ \ \ \  \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)\dfrac { 3 }{ 2.2 }  =\dfrac { 15 }{  11}  (3) \dfrac {  4}{ 9 } =\dfrac {  4}{ 9 } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \   (3)\dfrac { 4 }{ 2.8 }  =\dfrac { 10 }{ 7 }  \because   The ratio of  \dfrac {V  }{ I }   for device 1 is constant whereas for device 2 is not constant. 	herefore   Device 1 is ohmic and Device 2 is non-ohmic.

Question 3

If the resistance of a circuit is halved and the potential difference is kept constant, then the current will become

Accepted Answer

Answer is C.According to Ohm's law, V = IR. That is, I = V/R.Now, if the resistance is halved and the voltage remains the same, the new value of current becomes,  I'=\dfrac { V }{ \dfrac { R }{ 2 }  } =\dfrac { 2V }{ R } .\quad That\quad is,\quad I'=2(\dfrac { V }{ R } )=2I .Hence, the current will become double.

Question 4

In the given circuit the current  I_1  (magnitude only) is:

Accepted Answer

\dfrac {x-40}{40}+\dfrac {x-0}{30}+\dfrac {x-30}{40}=0 \Rightarrow \ \dfrac {x-40}{4}+\dfrac {x}{3}+\dfrac {x-30}{4}=0 \Rightarrow \ \dfrac {3x-120+4x+3x-90}{12}=0 \Rightarrow \ 10x=210 \Rightarrow \ x=21 	herefore I_1=\dfrac {40-x}{40}=\dfrac {40-21}{40}\equiv 0.4\ A

Question 5

If current  I_1=3Asin\omega t  and  I_2=4Acos\omega t , then  I_3  is

Accepted Answer

I_1=3A\sin\omega t I_2=4A\cos\omega t as from kirchoff's law I_3=I_1+I_2 I_2=3A\sin\omega t+4A\cos\omega t Multiplying and dividing the RHS with  \sqrt{(3A)^2+(4A)^2}=5A I_2=5A\left(\dfrac{3}{5}\sin\omega t+\dfrac{4}{5}\cos\omega t\right) as  \cos 53^0=\dfrac{3}{5}       \sin 53^0=\dfrac{4}{5} \Rightarrow I_3=5A(\cos 53^0\sin\omega t+\sin 53^0\cot\omega t) I_2=5A\sin(\cot+53^0) as  \sin(A+B)=\sin A\cos B+\cos A\sin B \boxed{I_3=5A\sin(\omega t+53^0)}

Question 6

Find the currents  I_1  and  I_2  for the circuit shown in given figure.

Accepted Answer

Consider node  a , using Kirchhoff's current law(using   +  for  I_{in}   -  for  I_{out} )                         \sum I_{at\, a\, node} = 0                         -3+2+I_{ab} = 0                          I_{ab} = 1 A Consider node  b ,                        I_{ba}+7+I_1 = 0            (  I_{ab} = -I_{ba} )                        -1-7 +I_1=0                          I_1 = -6A                      Consider node  d  ,                        I_{dc} +4-1 = 0                          I_{dc} = -3 A Consider node  c ,                          -I_2+6+I_{cd} = 0                           -I_2+6+3 = 0                           I_2 = 9A  option\, A\, is\, correct

Question 7

The equivalent resistance between the points x and y of the circuit. If each resistance is 25 ohm is

Accepted Answer

Simplified circuit diagram is drawn  As \dfrac{R_{1}}{R_{2}} = \dfrac{R_{3}}{R_{4}}      then   V_{BC} = 0  R_{XY} = \dfrac{50 	imes 50}{2 	imes 50}   R_{XY} = 25 \Omega

Question 8

In a potentiometer using two cells in series gives a balance length of  600cm . When the same cells are connected opposing each other then balance length is  100cm . The ratio of emf of larger to smaller cell is:

Accepted Answer

Case 1 \Rightarrow \left( { E }_{ 1 }+{ E }_{ 2 } ight) =I	imes r\left( 600 ight)  Case 2 \Rightarrow \left( { E }_{ 1 }-{ E }_{ 2 } ight) =I	imes r	imes100 \Rightarrow \dfrac { { E }_{ 1 }+{ E }_{ 2 } }{ { E }_{ 1 }-{ E }_{ 2 } } =\dfrac { 600 }{ 100 }  { E }_{ 1 }+{ E }_{ 2 }={ 6E }_{ 1 }-{ 6E }_{ 2 } 7{ E }_{ 2 }=5{ E }_{ 1 } \Rightarrow \dfrac { { E }_{ 1 } }{ { E }_{ 2 } } =7/5

Question 9

A potentiometer wire  AB  is  100\space cm  long and has a total resistance of  10\space ohm . If the galvanometer shows zero deflection at the position  C , then find the value of unknown resistance  R  in ohms

Accepted Answer

Given :      E = 10  volts             L = AB =100  cmThus potential difference A and C       V_{AC} = \dfrac{E}{AB}	imes AC = \dfrac{10}{100}	imes 40 = 4  volts 	herefore        V_{BG} = V_{PM} = V_{AC} =4   voltsUsing        V_{BG} = 5- Ir                 where  r =1\Omega 	herefore        4 = 5 -I(1)            \implies I = 1  AThus resistance         R = \dfrac{V_{PM}}{I} = \dfrac{4}{1} =4\Omega

Current Electricity