Recent content by rayhan619

so the resistance is 4 ohm and the voltage is 1.5 V so I = V/R = 1.5/40 = 0.375 A right?

oh ya. thanks

Next question, Find the current through 6.0 {Ohm resistor.

so area for cylinder is, A = 2 pi r ( r + h ). = 2(3.14)(2.5*10^-4){(2.5*10^-4)+0.1} m^2 = 1.57*10^-4 m^2 = 1.57 cm^2 so C = (1*10^-6 F)(1.57 cm^2) = 1.57*10^-6 F and V = -70 mV =...

V = IR I = V/R = 1.5 V/1 Ohm = 1.5 A so for 3 Ohm resistor voltage difference, V = IR = 1.5 A*3 Ohm = 4.5 V Right?

V = IR = 2A*4 Ohm = 8 V

there are like 6 parts of this problem. so for part a) C = Q/V V = 100 V C1 = 15*10^-6 F so Q = C1V = 100 V(15*10^-6 F) = 1.5*10^-3 C and how do I get the potential difference?

If i did it right then the total resistance should be 12 Ohm So I = V/R= 24 V/ 12 Ohm = 2 A. Right ? how do I get potential difference?

here is the pic

Homework Statement A real battery is not just an emf. We can model a real 1.5 battery as a 1.5 emf in series with a resistor known as the "internal resistance", as shown in the figure (Intro 1 figure) . A typical battery has 1.0 Ohm internal resistance due to imperfections that limit...

so the current through 4.0 ohm resistor, I = V/R = 24 V/4 Ohm = 6 A and the potential difference ?

I appreciate it.

Delta V = IR I = 500 microA = 500*10^-6 A Delta V = 5 V R = V/I = 5 V/(500*10^-6 A) = 1*10^4 Ohm Is this correct?

we know C = Q/V V = 100 V C1 = 15*10^-6 F so Q = C1V = 100 V(15*10^-6 F) = 1.5*10^-3 C C2 and C3 is in series. so 1/C = (1/C2 + 1/C3) = (1/30*10^-6)+(1/30*10^-6) = 66666.67 so C = 1.5*10^-5 would the Q be same after switching to point b? If yes, then we can figure out V using above equation.

Energy stored in a capacitor, U = 1/2(Q^2/C) = 1/2(Q/V) = 1/2 CV^2 If we know any two of Q,C, or V we can figure out the energy. C is 1 microF per square centimeters. we can get the area using A=(3.14)r^2 and get the C from there. and the V is given -70 mV. so we can use V and C to get U. right?