00:01
So for part a of this problem, we want to find the voltage that a voltmeter will read when connected at points a and b.
00:08
And since this is an ideal volt meter, we don't take into account its internal resistance.
00:14
So we only have to worry about adding the 18k and the 36k resistor.
00:20
And as you can see in the circuit there in series, so part a, the volt meter, the r equivalent is the 8k.
00:32
18k plus the 36k resistor.
00:37
So the total resistance is 54k oms.
00:46
And now since we're given the voltage of the battery, we can find the current of the circuit.
00:52
So current is equal to the voltage divided by the resistance.
00:56
So we're given the voltage of the battery as 48, divided by 54k.
01:05
So then our current is 8, 0 .8 .9 times 10 to the minus 4 amps...