00:01
Okay, so for problem 10, to find the current through the battery, the first thing we need to do is find the equivalent resistance.
00:08
So we're going to find the equivalent of the 7 and the 3, which are in parallel with each other by using the 1 over rule.
00:14
So 1 over that equivalent resistor is 1 over 7 plus 1 over 3, which if you combine those you get 10 over 21.
00:22
Then you take the reciprocal, you get 21 over 10, or 2 .1 ohms.
00:27
Now that equivalent resistance is in series with the 5 ohm resistor, and so we just add those together to find the total resistance of 7 .1 ohms.
00:38
And then we use that total resistance and the voltage from the battery in ohm's law to find the current.
00:46
So ohm's law, b equals i times r.
00:48
In other words, the current is the voltage from the battery over the total resistance of the circuit, and we get the current of 30 over 7 .1, or 4 .23 amps.
01:01
Now in question 11, we need to find the voltage drop across the 7 ohm resistor, the resistor number 2.
01:08
Now the voltage drop from here to here is going to be the same, so we're really finding both the voltage drop across r2 and r3, because they're going to be the same.
01:20
And it's going to be this 30 volts less the voltage drop from r1.
01:26
So we really need to find the voltage drop across r1, and then subtract that from 30.
01:31
And we can use ohm's law to do that.
01:33
So v1 is going to equal the current through the battery, because that's in series with the battery.
01:39
So the current we just found times r1, which is 5 ohms.
01:45
So we have our 4 .23 amps times 5 ohms, and we get 21 .1 volts dropped across r1.
01:59
And then if we take our 30 volts and subtract that, we'll get what v2 is...