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This is the second of our many quizzes on Direct and alternating current. Click on this to access more such competitive quizzes on Current Electricity and EMI.

Direct and alternating current Quiz 2

physics

An inductor blocks AC and offers a low resistance path for DC to pass through, and is used in a circuit as shown below where it is connected to an oscillating source. What is the inductor current when the inductance is connected to the oscillating source?<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/d15e2180-3e9b-4f6f-8847-59a735278cbd92944321591917342.png"/>

First determine the inductive reactance when connected to the power supply of given frequency and then arrive at the appropriate inductor current. 

  $\text{We are given}\,X_L = 240\Omega,f = 400Hz, f^{\prime} = 60Hz\,\text{and}\,\epsilon_{rms} = 240V.\\\text{In general, the inductive reactance is given as:}\\ X_{L} = \omega L = 2\pi f L\\\Rightarrow X_L \propto f\\\text{Let}\,X_{L}^{\prime}\,\text{be the reactance of the same inductance at 60Hz.}\\\Rightarrow \dfrac{X_L^{\prime}}{X_L} = \dfrac{f^{\prime}}{f}\\\Rightarrow X_L^{\prime} = \dfrac{f^{\prime}}{f} \times X_L \\ \Rightarrow X_L^{\prime} = \dfrac{60}{400} \times 240 \\ \Rightarrow X_L^{\prime} = 36\Omega\\\text{Let}\,I_{rms}\,\text{be the current that flows through the inductance when connected to the power supply.}\\I_{rms} = \dfrac{\epsilon_{rms}}{X_{L}^{\prime}} \\ \Rightarrow I_{rms} = \dfrac{240}{36} \\ \therefore I_{rms} = 6.67A$

A power station has to transmit $10,000\,W$. It can be done either by sending $100\,A$ at $100\,V$ or $10\,A$ at $1000\,V$. If the resistance of the transmission line is $0.25\,\Omega$ then calculate the power that the generator will have to supply in order to account for the power losses in each case. Which transformer is needed to convert the lower voltage to higher voltage? 

 First find the power loss in each case, and the power transmitted by the station should account for this loss in addition to the actual load being transmitted.  

  $\text{Given the resistance of the transmission line as}\,R = 0.25\Omega\\\text{When}\,I = 100A\,\text{the power loss}\, P = I^2R = 100^2 \times 0.25 = 2500W.\\\text{Power to be supplied by the generator}\,P_{supply} = 10,000+2500 = 12,500W\\\text{When}\,I = 10A\,\text{the power loss}\,P = I^2R = 10^2 \times 0.25 = 25W\\\text{Power to be supplied by the generator}\,P_{supply} = 10,000+25 = 10,025W\\\text{We have primary emf}\,\epsilon_{p} = 100V\,\text{and secondary emf}\,\epsilon_{s} = 1000V\\\text{The ratio of number turns can be given as:}\\\dfrac{N_s}{N_p} = \dfrac{\epsilon_{s}}{\epsilon_{p}} \\ \Rightarrow\dfrac{N_s}{N_p}= \dfrac{1000}{100} \\\Rightarrow\dfrac{N_s}{N_p}= 10\\\therefore N_{s} = 10N_{p}\\\text{This is a step-up transformer,}\text{where the number of coils in the secondary is greater than the number of coils in the primary.}$ 

The energy $U$ of an $LC$ circuit continuously oscillates between electric and magnetic energies, but the total energy remains constant. What is the total energy of this circuit with capacitance $C$, inductance $L$, peak charge $Q_0$ and angular frequency $\omega$? 

 Use instantaneous charge expression to find instantaneous current and arrive at the electric and magnetic energies to determine total energy.

$\text{The instantaneous charge in the circuit can be given as}\,Q= Q_0cos(\omega t).\\\text{We know that instantaneous current oscillates as:}\\I = \dfrac{dQ}{dt} \\ \Rightarrow I= \dfrac{d}{dt} Q_0cos(\omega t) \\ \Rightarrow I= -\omega Q_0sin(\omega t)\\\text{We are given that the energy oscillates between electric and magnetic energies.}\\\text{The electrical energy arises by virtue of the capacitor, i.e.,}\\U_{E} = \dfrac{Q^2}{2C} \\ \Rightarrow U_{E}= \dfrac{ Q_0^2}{2C}cos^2(\omega t)\\\text{The magnetic energy arises by virtue of the inductor, i.e.,}\\U_{B} = \dfrac{1}{2}LI^2 \\\Rightarrow U_{B} = \dfrac{1}{2}L (-\omega Q_0sin(\omega t))^2 \\ \Rightarrow U_{B} = \dfrac{1}{2}L\omega^2Q_0^2sin^2(\omega t)\\\text{Plugging in}\, \omega = \dfrac{1}{\sqrt{LC}}\\U_B = \dfrac{Q_0^2}{2C}sin^2(\omega t)\\\text{Therefore, the total energy:}\\ U = U_E +U_B \\ \Rightarrow U = \dfrac{ Q_0^2}{2C}cos^2(\omega t) + \dfrac{Q_0^2}{2C}sin^2(\omega t)\\ \Rightarrow U =  \dfrac{Q_0^2}{2C}(cos^2(\omega t) +sin^2(\omega t)) \\ \therefore U = \dfrac{Q_0^2}{2C}$

Find the magnitude of the peak supply voltage of the following series $LCR$ circuit with angular frequency $\omega = 377\,Hz$ and peak current $I_{0}= 0.255\,A$.<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/f732bb5f-9f04-4de5-863c-a61668fcd5542099102216709250099.png"/> 

Find the impedance of the LCR circuit and its relation with peak current to find the supply voltage.  

 $\text{Given}\,\omega = 377Hz\,\text{peak current}\,I_{0} = 0.255A.\\\text{From the circuit diagram we have}\,R = 200\Omega,L = 0.6H\,\text{and}\,C = 3.5 \times 10^{-6}F\\\text{We are supposed to find the peak voltage}\,\epsilon_0.\\\text{Now, the inductive reactance is given by,}\\ X_L = \omega L \\\Rightarrow X_L = 377 \times 0.6 \\ \Rightarrow X_L = 226\Omega\\\text{The capacitive reactance is given by:}\\ X_C = \dfrac{1}{\omega C} \\ \Rightarrow X_C = \dfrac{1}{377 \times 3.5 \times 10^{-6}} \\\Rightarrow X_C = 758\Omega\\\text{Therefore, the impedance of the LCR circuit will be:}\\Z= \sqrt{R^2+(X_L-X_C)^2} \\ \Rightarrow Z = \sqrt{250^2 + (226-758)^2} \\ \Rightarrow Z = 588\Omega\\\text{We know that the peak voltage is given as:}\\\epsilon_0 = I_0 Z \\\Rightarrow\epsilon_0 = 0.255 \times 588 \\ \therefore\epsilon_0 = 149.94 \approx 150V$

 From the following figure, determine the resonance frequency if the amplitude of current in an $RLC$ circuit is given by $I_0 = \dfrac{\epsilon_0}{\sqrt{R^2+\left(\omega L-\dfrac{1}{\omega C}\right)^2}}$.<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/7ca40ef1-0f28-4d7d-866d-80ed7976299b7734456448099269939.png"/> 

Arithmetically solve the $RLC$ current and the current at the upper and lower power frequencies, and rearrange to arrive at the required relation.

 $\text{From the figure, at}\,\omega_1\,\text{and}\,\omega_2\,\text{which are the lower and upper power frequencies, the current amplitude is given by:}\\I_0 = \dfrac{1}{\sqrt{2}}\dfrac{\epsilon_0}{R}\\\Rightarrow \dfrac{\epsilon_0}{\sqrt{R^2+\left(\omega L-\dfrac{1}{\omega C}\right)^2}} =  \dfrac{1}{\sqrt{2}}\dfrac{\epsilon_0}{R}\\\text{Inverting the entire equation and squaring both sides we get:}\\\Rightarrow R^2 + \left(\omega L - \dfrac{1}{\omega C}\right)^2 = 2R^2\\\Rightarrow \left(\omega L - \dfrac{1}{\omega C}\right)^2 = R^2\\\Rightarrow \left(\omega L - \dfrac{1}{\omega C}\right) = \pm R\\\Rightarrow \omega_1 L - \dfrac{1}{\omega_1 C} = -R\,\text{and}\, \omega_2 L - \dfrac{1}{\omega_2 C} = +R\\\text{Adding the above two equations we get:}\\\Rightarrow (\omega_1+\omega_2)L = \dfrac{1}{\omega_1 C} + \dfrac{1}{\omega_2 C} \\\Rightarrow (\omega_1+\omega_2)L= \dfrac{(\omega_1 +\omega_2)}{(\omega_1 \omega_2)C}\\\Rightarrow \omega_1 \omega_2 = \dfrac{1}{LC}\\\text{But we know that angular resonance frequency}\,\omega_r = \dfrac{1}{\sqrt{LC}}\\\Rightarrow\omega_1 \omega_2 = \omega_r^2 \\ \therefore \omega_r = \sqrt{\omega_1 \omega_2}$