00:01
Part a of the given problem, the energy initially stored in the capacitor is given by u .n.
00:07
Is equal to q0 squared divided by two times of capacitance.
00:11
So i'm studying a values for q0, which is 0 .0069 coulum square divided by 2 times of 4 .62 times 10 to the power minus 6 ferrard.
00:24
This gives us 5 .15 joules.
00:29
Part b of the problem, we need to find the electrical power dissipated in resistor just after the connection is made, which is given by p -k -nod is equal to i -nod squared times r, which is equal to q -nod square divided by rc, which is i -nod square, so square is outside, sorry, times r, we substitute the values for q -0, so 0 .0 .0.
01:01
Square divided by 850 ooms times capacitors of 4 .62 times 10 power minus 6 ferrad this whole square this square this gives us power of 262 20 watts part c of the problem part c in the resistor at the instant when the energy stored in the capacitor has decreased with a half value as calculated in part a.
01:43
The charge or the capacitor at any instant is given by q is equal to q nod, e exponential minus t divided by rc.
01:54
So there are no 2 minus 1 minus sign here.
01:58
The energy store is therefore converted as u is equal to q squared divided by by 2c.
02:04
Substruiting value of q, which is q0, square, divided by 2c times exponential minus t over rc, this whole square...