Free Study Material

NCERT Solutions

NCERT Books

Revision Notes

Practice Problems

This is one of our many quizzes on Resistance. Click on this to access more such competitive quizzes on Current Electricity.

Resistance Quiz 2

physics

Calculate the equivalent resistance of the network shown in the figure between A and C? $ \\ $ <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/ea9c66e9-3de7-4a02-879e-a2652fd7e10c7994937496648465146.png"/>

 Recall the concept of Wheatstone bridge and eliminate the resistance which is of no use. Further, determine the resistances in series and parallel to calculate the equivalent resistance. 

From the diagram, the two leftmost and the two rightmost resistances are in series. And since it's the case of a balanced Wheatstone bridge, the middle resistance is of no use. Therefore, the effective resistance can be calculated as - ${{R}_{eq}}=\left[ R+2R \right]\parallel \left[ 2R+4R \right]=\left[ 3R\parallel 6R \right] \\ \therefore {{R}_{eq}}=\dfrac{3R\times 6R}{(3R+6R)}=2R$

 Which of the following statements is incorrect?

 Resistance of a conductor depends upon the ion density, length, area, temperature and various other factors. But the specific resistance or the resistivity of a conductor only depends upon the nature and temperature of the material.

  Specific resistance or resistivity of a conductor only depends upon the nature and temperature of the material. The resistance of a conductor is directly proportional to the ion density of the conductor. Also, changing the arrangement of the atoms will definitely cause change in resistance as in tetrahedral geometry, there would be 4 atoms whereas in octahedral geometry there are $6$ atoms per block.

For the given graph, which of the following statements is true? ‘$I$’ is current, ‘$V$’ is voltage and ${{T}_{1}}$ and ${{T}_{2}}$ are temperatures. <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/15682d0c-ccfb-41a8-b6c2-aeee01edb6dc3524089384450995193.png"/>

 The resistance of a resistor increases with increase in temperature. And from the given graph is a $V-I$ graph, the slope is directly proportional to the inverse of resistance, that is, conductance. 

The slope of the given curve $\dfrac{I}{V}=\dfrac{1}{R}$ And $\dfrac{1}{R}$ conductance of the wire. Therefore, Slope is directly proportional to the conductance. The slope of the graph at temperature ${{T}_{1}}$ is greater than that at temperature ${{T}_{2}}$. Since resistance at ${{T}_{2}}$ is $\rangle $ resistance at ${{T}_{1}}$. $\therefore {{T}_{2}} > {{T}_{1}}$

 Among the shown circuits, for which circuit the loss of energy of each component is equal? Consider that the same voltages are applied to each circuit. 

 Try to relate the proportionalities of heat loss, voltage and resistance. The resistances connected in parallel should be equivalent for fixed voltage for equivalent heat loss. 

For equal heat loss, the resistances connected in parallel should be equal for constant voltage. Secondly, the resistance connected in series with the parallel resistance should be ${{\dfrac{1}{4}}^{th}}$ the value of the parallel resistances. Energy is given by, $E=VIt \\ \Rightarrow E=\dfrac{{{V}^{2}}t}{R} $ Voltage drop will be different in series connection and will be equal in parallel connection. So, resistance must be equal in parallel as $E\propto \dfrac{1}{R}$. So, the correct option will be either (a) or (b). Now, for the series resistor to have the same voltage drop, the resistance connected in series with the parallel resistance should be ${{\dfrac{1}{4}}^{th}}$ the value of the parallel resistances. Therefore, option (a) will be the correct answer.

In the below shown circuit diagram, the resistance of each resistor is $R$. The value of each resistance at temperature ${{20}^{o}}C$ was found to be $3.75\Omega $. If the temperature coefficient of each resistance is $0.0025\,pe{{r}^{o}}C$, then the current passing through the circuit at ${{100}^{o}}C$ is? <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/0242891a-4e25-4a6d-9d8b-89edca87cf9e2678596411698613664.png"/>

 Recall the concepts of variation of resistance with temperature and find the resistance at the given temperature. Calculate the equivalent resistance and use Ohm's law to find current.

Let the resistance of each resistance at temperature ${{100}^{o}}C$ be ${{R}_{2}}$ Therefore, by using the concept of variation of temperature, $\alpha =\dfrac{{{R}_{2}}-{{R}_{1}}}{{{R}_{1}}{{t}_{2}}-{{R}_{2}}{{t}_{1}}}=\dfrac{{{R}_{2}}-3.15}{(3.15\times 100)-({{R}_{2}}\times 20)}=0.0025 \\ \Rightarrow {{R}_{2}}=3.75$ Hence, the required resistance is $3.75\text{ }\Omega $. Where ${{R}_{1}}$ is the resistance of each resistor at ${{20}^{o}}C$ and $\alpha $ is the temperature coefficient of resistance. Now, the equivalent resistance of the circuit at ${{100}^{o}}C$ is $0.8$. By using ohm's law, $I=\dfrac{V}{R}=\dfrac{12}{0.8}=15\,A$