# Tutor profile: Anurag S.

## Questions

### Subject: Physics (Electricity and Magnetism)

A Capacitor initially having a charge $$20\mu C$$ is connected across an ideal Inductor (initially at rest). The Capacitor has a capacitance of $$4\mu F$$ and the Inductor has an inductance of $$100mH$$. Calculate the maximum oscillating current in the circuit.

The Capacitor has an initial charge, therefore, a voltage across its terminals. $$ChargeStored = (Capacitance)*(Voltage)$$ $$Q=CV$$ $$V=\frac{Q}{C}=\frac{20\mu C}{4\mu F}$$ $$V_{cap}=5V$$ So, a voltage of 5V is across the Capacitor's plate before connecting it to the Inductor. Energy Stored in the Capacitor- $$E_{cap}=\frac{1}{2}CV_{cap}^{2}$$ $$E_{cap}=\frac{1}{2}(4\mu F)(5V)^{2}=50\mu J$$ This Energy oscillated between the Inductor's magnetic field and Capacitor's Electric Field rendering the Inductor's current and Capacitor's voltage to be sinusoidal (altering between positive and negative peaks). Also, $$E_{ind} = \frac{1}{2}LI_{ind}^2$$ Now we get the following- $$V_{cap} = V_{cap}(max)\longrightarrow E_{cap} = 50\mu J\;and\;E_{ind} = 0$$ $$I_{ind} = I_{ind}(max)\longrightarrow E_{ind}= 50\mu J\;and\;E_{cap}=0$$ Therefore, for maximum current in Inductor will occur at the instant when Inductor's Stored Energy $$E_{ind}$$ is maximum (equal to initial starting value of total System's Energy). $$E_{ind}=\frac{1}{2}LI_{ind}^2$$ or, $$I_{ind}(max) =\sqrt{\frac{2E_{ind}(max)}{L}}$$ $$I_{ind}(max) =\sqrt{\frac{2*50\mu J}{100mH}}$$ $$I_{ind}(max) =31.62mA$$

### Subject: Electrical Engineering

An ideal transformer having 90 turns on the primary and 2250 turns on the secondary is connected to a 200V, 50Hz source. The Load across the secondary draws a current of 2A at a power factor of 0.8 lagging. Find: (a) The effective value of the Secondary Voltage and Primary Current (b) The instantaneous current in the primary when the instantaneous current in the secondary is 100mA. (c) The peak flux linked by the secondary.

Let the primary side voltage and current be $$V_{1}$$ and $$I_{1}$$ respectively, and, corresponding values on Secondary side be $$V_{2}$$ and $$I_{2}$$. Then, for an Ideal Transformer we have, $${\frac{V_{1}}{N_{1}}}={\frac{V_{2}}{N_{2}}}$$ {Voltage-per-turn is equal on both sides} $$N_{1}I_{1}=N_{2}I_{2}$$ {Ampere-turns on both sides is equal} NOTE-The above relations are true for RMS(effective) values as well as Instantaneous values of Voltage and Current in Ideal Transformer. Now proceeding with the question. (a) Finding effective Secondary Voltage ($$V_{2(rms)}$$): Using first relation we get $$V_{2(rms)} = \frac{V_{1(rms)}}{N_{1}}N_{2}$$ $$V_{2(rms)} = \frac{200V}{90}*2250=5000V$$ Finding effective Primary current ($$I_{1(rms)}$$): using second relation $$I_{1(rms)}=\frac{N_{2}I_{2(rms)}}{N_{1}}$$ $$I_{1(rms)}=\frac{2250*2A}{90}=50A$$ (b) Since given relations are also true for instantaneous values of Voltage and Current, therefore- $$I_{1}=\frac{N_{2}I_{2}}{N_{1}}$$ $$I_{1}=\frac{2250*100mA}{90}=2500mA=2.5A$$ NOTE- When Voltage is stepped up (or increased) then Current is stepped down (or decreased) and vice versa, when traveling from one side of the transformer to other. But, the Apparent power (product of Voltage and Current) remains the same on both sides i.e., $$V_{1}I_{1}=V_{2}I_{2}$$ (c) The induced EMF in transformer coils is given as- $$E_{1(rms)}=4.44f\phi_{max}N_{1]$$ {derived using Faraday's Law of Electro-magnetism : $$V=\frac{Nd\phi}{dt}$$} In ideal transformer $$V_{1}=E_{1}$$, therefore, $$\phi_{max}=\frac{E_{1(rms)}}{4.44fN_{1}}$$ $$\phi_{max}=\frac{200}{4.44*50*90}=10mWb$$ The Ideal Transformer is almost an accurate model for analysing a Practical Transformer because the losses are minimal in this electrical device with Efficiencies usually exceeding 99%.

### Subject: Algebra

Find the Remainder when $$x^{3}-ax^{2}+6x-6a$$ is divided by $$x-a$$.

Given polynomial is of degree 3 (often called cubic polynomial). When we divide say $$p(x)=x^{3}-ax^{2}+6x-6a$$ (Dividend) by $$x-a$$ (Divisor) then by division rule i.e., $$Dividend = (Divisor)*(Quotient) + Remainder$$ we get something like this: $$p(x) = (x-a)(polynomial-of-degree-2)+Remainder$$ Notice that the degree on RHS still adds to a total of 3 (degree 1 from Divisor and degree 2 from Quotient) as on LHS, since, the sum of the degree of individual multiplicative factors of a polynomial is equal to the total degree of that polynomial. Now if we put $$x=a$$ in the above equation then we get $$p(x=a)=(a-a)(polynomial-of-degree-2)+Remainder$$ which simplifies to, $$p(x=a) = Remainder$$ which tells us that if we divide a polynomial $$p(x)$$ by $$x-a$$ then the Remainder is same as $$p(x=a)$$. This result is called Factor Theorem. In General, we can state that- If we have a polynomial of $$p(x)$$ of degee $$n$$ which is divided by $$(x-a)$$ then $$p(x) = (x-a)*(polynomial-of-degree-[n-1])+Remainder$$ and the Remainder is equal to $$p(x=a)$$. So, answer to our question is $$Remainder = p(x=a) = a^{3}-a*a^{2}+6a-6a = 0$$. Therefore $$Remainder = 0$$ and $$x-a$$ is a factor of given polynomial $$p(x)$$ i.e., $$x-a$$ compeletely divides (remainder = 0) our cubic polynomial $$p(x)=x^{3}-ax^{2}+6x-6a$$.

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