00:01
For part a of the given problem, the resonance frequency, omega -r, is given by 1 divided by square root of lc, and the resonance frequency r is simply 1 divided by 2 -py, square root of lc.
00:23
For omega -r resonance, we'll simply substitute the given values that is, is 1 times 10 to the power minus 2 times 1 times 10 to power minus 5 in the square root this will give us omega r to be 3 .16 times 10 to the power 3 radiance per second where far similarly we'll just multiply a 2 pi in the denominator and if r we get here is 5 .03 times 10 to the power 2 power 2 this was for part a.
01:10
For part b of the problem, we are asked to find the voltage across resistance, vr, which is given by ir.
01:21
And i here is p -0 divided by r, which is 1.
01:28
So we'll multiply 1, multiply with 10.
01:32
This will give us 10 volts.
01:35
For a vc, voltage across capacitor is i times xc, and this we can write as i divided by omega c times c...