00:01
So here we have a capacitor that charges up to a voltage of 35 volts.
00:08
And this capacitor sends this voltage every 1 over 60 seconds, which is our t.
00:15
And we're also told that this charge must hold up until it reaches 1 volt.
00:20
And we also know that the current running through here is given as i is equal to 1 .2 amps.
00:30
So for part a, to find the resistance, we simply use the equation, which is a variation of oms law, as r is equal to the voltage divided by the current.
00:43
The voltage here is 35, that's what being supplied by the battery, divided by the current, which is just 1 .2.
00:53
So for part a, the resistance is r equal to 29 .2 oms.
01:04
Now for part b to find the capacitance c, we need to use the equation for the voltage in a discharging capacitor...