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

Electromotive force Quiz 2

physics

The resistances of three conductors vary with time as shown in the $R-t$ time graph where $R$ is resistance and $t$ is the time. Determine the inequality relation between the emf’s across each resistance. <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/ed88f34e-e792-4fdc-9056-d68ab9b04f258906514041141030005.png"/>

 The emf across these resistors will be equal to the potential difference across the resistors. Thus, emf will be the product of the constant current and the time varying resistance. 

The emf of a cell is the driving force of electrons in a circuit. In other words, the emf is actually the potential difference across the circuit. Let the three resistors ${{R}_{1}}$ , ${{R}_{2}}$ and ${{R}_{3}}$ form three different circuits such that the three resistors are connected to three different cells respectively. All three circuits are independent of each other and there is no connection between any two circuits. Let the emf’s of these three cells be ${{E}_{1}}$ , ${{E}_{2}}$ and ${{E}_{3}}$ respectively. Given that the current flowing through all the resistors is equal and it is always constant, let the value of current be $i$. So, the emf’s are given by; ${{E}_{1}}=i{{R}_{1}}$ ${{E}_{2}}=i{{R}_{2}}$ ${{E}_{3}}=i{{R}_{3}}$ From the graph it is given that; ${{R}_{1}}>{{R}_{2}}>{{R}_{3}}$ for $t\in \left[ 0,12 \right)$ And ${{R}_{1}}>{{R}_{3}}>{{R}_{2}}$ for $t\in \left( 12,\infty \right)$ But emf is directly proportional to resistance; ${{E}_{1}}>{{E}_{2}}>{{E}_{3}}$ for $t\in \left[ 0,12 \right)$ and ${{E}_{1}}>{{E}_{3}}>{{E}_{2}}$ for $t\in \left( 12,\infty \right)$

 Determine the acceleration of each electron in a conductor if the resistance of the conductor is $6\text{ }\Omega $ and $2\text{ C}$ of charge flows through it in $\text{4 sec}$. 

 The electrons move because of the electromotive force. This electromotive force is directly proportional to the acceleration of the electrons. 

The current flowing through the conductor is given as; $I=\dfrac{Q}{t}=\dfrac{2}{4}=\dfrac{1}{2}\text{ A}$ The emf of this conductor is given by; $ E=IR \\ \Rightarrow E=\dfrac{1}{2}\times 6 \\ \Rightarrow E=3\text{ V} $ The force acting on an electron is given by; $ F=eV \\ \Rightarrow {{m}_{e}}a=eV \\ \Rightarrow \left( 9.1\times {{10}^{-31}} \right)a=\left( 1.6\times {{10}^{-19}} \right)\times 3 \\ \Rightarrow a\left( 9.1\times {{10}^{-31}} \right)=4.8\times {{10}^{-19}} \\ \Rightarrow a=\dfrac{4.8\times {{10}^{-19}}}{9.1\times {{10}^{-31}}} \\ \therefore a=5.3\times {{10}^{11}}\text{ m}{{\text{s}}^{-2}}$

 Electrons are flowing through a circular circuit such that each electron traces a circle of radius $R$. If the radial acceleration of each electron is $a\left( t \right)=2{{t}^{2}}$ calculate the emf for the circular conductor. 

 The emf of the circular conductor is directly proportional to the electric force acting on one electron.

The radial acceleration of an electron is given as; $ \dfrac{{{v}^{2}}}{R}=a\left( t \right) \\ \Rightarrow \dfrac{{{v}^{2}}}{R}=2{{t}^{2}} \\ \Rightarrow v=t\sqrt{2R} $ The tangential acceleration of the electron is given by; ${{a}_{t}}=\dfrac{dv}{dt} \\ \Rightarrow {{a}_{t}}=\sqrt{2R}$ The electric force acting on the electron is given by; ${{m}_{e}}{{a}_{t}}=eV \\ \Rightarrow \left( 9.1\times {{10}^{-31}} \right)\sqrt{2R}=\left( 1.6\times {{10}^{-19}} \right)V \\ \Rightarrow V=\dfrac{9.1\times {{10}^{-31}}}{1.6\times {{10}^{-19}}}\sqrt{2R} \\ \therefore V=5.69\times {{10}^{-12}}\sqrt{2R}$

 Consider an ideal case where an electron is moving through a conductor such that the velocity of this electron is given by $v\left( t \right)=2{{t}^{2}}+4t$ . Calculate the power dissipated at $t=6\text{ sec}$ if the resistance of the conductor is $R$ . 

 The electron is moving through the conductor because there is an electromotive force in the conductor, this electromotive force (emf) is proportional to the acceleration of the electron. Power is a ratio of the square of emf and the resistance of the conductor. 

The electric force acting on the electron is given by; ${{F}_{e}}={{m}_{e}}a=qV$ Acceleration is given by; $a=\dfrac{dv}{dt} \\ \Rightarrow a=4t+4 $ Substituting the values; $\left( 9.1\times {{10}^{-31}} \right)\left( 4t+4 \right)=\left( 1.6\times {{10}^{-19}} \right)V \\ \Rightarrow V=\dfrac{\left( 9.1\times {{10}^{-31}} \right)}{1.6\times {{10}^{-19}}}\left( 4t+4 \right) \\ \Rightarrow V=5.69\times {{10}^{-12}}\left( 4t+4 \right)$ The power dissipated is given by; $P=\dfrac{{{V}^{2}}}{R} \\ \Rightarrow P\left( t \right)=\dfrac{{{\left( 5.69\times {{10}^{-12}} \right)}^{2}}{{\left( 4t+4 \right)}^{2}}}{R} \\ \Rightarrow P\left( 6 \right)=\dfrac{32.38\times {{10}^{-24}}\times 784}{R} \\ \therefore P\left( 6 \right)=\dfrac{2.5\times {{10}^{-21}}}{R}$

 Calculate the power dissipated when a certain amount of charge flows through a conductor of volume $24\text{ }{{\text{m}}^{3}}$ for $4\text{ sec}$ . Assume that electrons have an acceleration of $2\times {{10}^{12}}\text{ m}{{\text{s}}^{-2}}$ and the electron density is $\rho $ electrons per cubic meter.

 The electron density per cubic meter is given in the question. Thus, the total charge flowing can be calculated which in turn will give the current flowing for the required time. 

The electric force acting on an electron is given by; ${{m}_{e}}a=eV \\ \Rightarrow 9.1\times {{10}^{-31}}\times 2\times {{10}^{12}}=1.6\times {{10}^{-19}}V \\ \Rightarrow V=\dfrac{9.1\times {{10}^{-31}}}{1.6\times {{10}^{-19}}}\times 2\times {{10}^{12}} \\ \Rightarrow V=5.69\times {{10}^{-12}}\times 2\times {{10}^{12}} \\ \Rightarrow V=11.38\text{ V} $ The electrons are flowing for $4\text{ sec}$ then $I=\dfrac{Q}{t}$ The charge is given by; $Q=ne=24\rho e$ Thus, the current will be; $I=\dfrac{24\rho e}{4}=6\rho e\text{ A}$ The power dissipated is given by; $P=VI \\ \Rightarrow P=11.38\times 6\rho e \\ \therefore P=68.28\rho e\text{ W}$