ResistorA of 10 Z Ohms ResistorB of 20Z Ohms and a 10Volt w

Resistor-A of (10 + Z) Ohms), Resistor-B of (20+Z) Ohms, and a 10-Volt was connected in series. The connection between resistor-A and resistor-B was grounded. Find the voltage at the two terminals of each resistor. A 2 Mega-Ohms resistor was connected in series to a parallel plate capacitor with a gap of (50 + Z) micron and area = 2 square meters and thickness (10 + Z) cm. The circuit battery of 10 Volt was grounded such that one plate was grounded. Superman crushed the positively charged plate and a sphere was formed. Find the new time constant. R_A = 10 + 95 = 105 ohms. R_B = 20 + 95 = 115 ohms.

Solution

Current in the circuit = 10/(10+ Z + 20+Z)

= 10/(30+2Z)

= 10/(30+2*95)

= 0.0455 A

So, voltage at the terminal of resistor A = 0.0455*(10+95)

= 4.78 V

Voltage at terminal of resistor B = 0.0455*(20+95)

= -5.23 V

B)

Volume of the plate = 2*(10+95)*10^-3

= 0.21 m3

So, with the radius of the sphere as r,

(4/3)*pi*r^3 = 0.21

So, r = 0.369 m3

So, capacitance of the spherical capacitor ,

C = 4*pi*8.854*10^-12*0.369

= 4.11*10^-11 F

So, time constant T = R*C

= 2*10^6*4.11*10^-11

= 8.22*10^-5 s