Messages posted by karthik2007

-5. Negative sign for the 180 deg phase shift

Apply a horizontal force F. Then F x (distance between F and corner of toppling) = N (distance to COM) is the limiting case

The resistance seen by the capacitor is R || 2R which is 2R/3. Now find the time constant, which is 2RC/3

What he said is correct. Apart from that, you can also look at this way. If two surfaces intersect, then a charge will have to move along both paths simultaneously (0 work is needed for it to move), which is impossible.

The time constant is given by CR, where C is the capacitance, and R is the resistance connected in series with the capacitor.

At 20 deg C, the resistance is 15/18.5

Hence

Where t = 293 (20 + 173)

Similarly, 15/17.2 = Ro(1+a (92+273))

Two equations, two unknowns. Solve for Ro and alpha. Then use Ro = PL/A to find out the resistivity P.

The temperature coefficient of the material used is needed to solve the problem.

A capacitor discharges the charge stored in it through a resistor. It does so in an exponentially decaying fashion.

The above equation gives the current flowing through the resistor at any time t. The net charge passing through the resistor is given by

Integrate the above expression to get your answer.Note that CR is the time constant of an RC circuit. It is the amount of time taken for the charge to deprecate to 1/e times it's maximum value.

Use conservation of energy.

Where theta = 20, h = 75.

u = initial velocity given, and gamma = 1/2.

Hence find out v.

For domestic applications (household wiring), the voltage is of importance. Devices will work if a particular voltage is applied across their terminals (230.94V). The current is not of importance here (Though starting current is required for a few devices like Acs, fans). So in such cases, we can regard the Voltage to remain constant for power calculations.

Straightforward. Use Y = FL/Al

If you consider the bottom edge, then friction doesn't come into play.You have the following equations for the limiting case :F(3a/4) = N(a/2)N=mg.Solve to get F = 2mg/3.

Quite simple.

ma = kx+mg.

Get a, then use the equations of motion to get your time.

A transformer will have maximum efficiency when its constant losses = variable losses.

Copper losses at nth load (n<1) = 1600n² (As copper losses are directly proportional to the square of the current, and the current at nth load = I_fullload x n)

Thus, 900 = 1600n²

n = 3/4 = 75%.

Thus the efficiency is maximum at 75% load.

Yup. What the above expert said is perfectly right. The core losses are generally of the order of 100W, and the full load current is 7.5 A (generally), with the winding resistance being 2 ohms or so (varies from transformer to transformer)

1/alpha, where alpha is the thermal coefficient of the resistance.

What is point P? be more clear.

I missed a cos @ term in the integrand. Here cos @ = p/sqrt(p^2 + a^2).
Add that and you'll get the right answer.

@above - there is a p term in the integrand if you didn't see the first step, so the integral will be ln and not arc tan.

The given surface is a loop (circular) at a height a from the origin. Consider an elemental area by incrementing the radius and the angle it subtends by small amounts. So the elemental area becomes $\rho d\rho d\phi$

The flux through this elemental area = $\frac{\mu_o t}{4\pi (a^2+\rho ^2)} \rho d\rho d\phi$

The net flux through the loop is $\int^{\rho}_0 \frac{\mu_o t}{4\pi (a^2+\rho ^2)} \, d \rho \int^{2\pi}_0 \, d\phi$

Thus, Net flux = $\frac{\mu_o t}{4} (ln(2a^2) - ln(a^2))$ = $\frac{\mu_o t}{4} \ln 2$

Now, the emf induced = $\frac{d(flux)}{dt} = \frac{\mu_o \ln 2}{4}$

diagram is needed for this one man.

Preparing for IIT-JEE ?

Arihant Revision Package for IIT JEE - Books, Practice Tests + Rank Predictor

@ INR 1,995/-

For Quick Info