# Tutor profile: Ajit P.

## Questions

### Subject: Physics (Electricity and Magnetism)

Why is the starting current high in a DC motor?

In DC motors, Voltage equation is V=Eb-IaRa (V = Terminal voltage, Eb = Back emf in Motor,Ia = Armature current, Ra = Armature resistance). At starting, Eb is zero. Therefore, V=IaRa, Ia = V/Ra , where Ra is very less like 0.01ohm.i.e, Ia will become enormously increased.

### Subject: Basic Math

A sheet of metal 12 inches by 10 inches is to be used to make a open box. Squares of equal sides x are cut out of each corner then the sides are folded to make the box. Find the value of x that makes the volume maximum.

We first use the formula of the volume of a rectangular box. V = L * W * H The box to be made has the following dimensions: L = 12 - 2x W = 10 - 2x H = x We now write the volume of the box to ba made as follows: V(x) = x (12 - 2x) (10 - 2x) = 4x (6 - x) (5 - x) = 4x (x 2 -11 x + 30) We now determine the domain of function V(x). All dimensions of the box must be positive or zero, hence the conditions x > = 0 and 6 - x > = 0 and 5 - x > = 0 Solve the above inequalities and find the intersection, hence the domain of function V(x) 0 < = x < = 5 Let us now find the first derivative of V(x) using its last expression. dV / dx = 4 [ (x 2 -11 x + 3) + x (2x - 11) ] = 3 x 2 -22 x + 30 Let us now find all values of x that makes dV / dx = 0 by solving the quadratic equation 3 x 2 -22 x + 30 = 0 Two values make dV / dx = 0: x = 5.52 and x = 1.81, rounded to one decimal place. x = 5.52 is outside the domain and is therefore rejected. Let us now examine the values of V(x) at x = 1.81 and the endpoints of the domain. V(0) = 0 , v(5) = 0 and V(1.81) = 96.77 (rounded to two decimal places) So V(x) is maximum for x = 1.81 inches.

### Subject: Electrical Engineering

What happens if two charged capacitors are connected in parallel. What will be the final voltage of the parallel combination?

Let us assume the two capacitors are C1 and C2 charged to voltages V1 and V2 respectively. Charge on C1 : Q1 = C1*V1 Charge on C2: Q2 = C2*V2 After connecting them in parallel, the the voltage across the capacitors are same and that is the voltage across the junction point. Let us name that voltage as V. For simplicity, is we consider the capacitors to be ideal, the charge is always conserved. that means the total final charge Q = Q1 + Q2 Final capacitance C = C1 + C2 (parallel connection ) Thus final voltage V = total final charge/final capacitance = (Q1 + Q2)/ (C1 + C2) = (C1*V1 + C2*V2)/ (C1 + C2)

## Contact tutor

needs and Ajit will reply soon.