In the circuit shown in Fig. E26.33 all meters are idealized and the batteries have no appreciable internal resistance.

(a) Find the reading of the voltmeter with the switch S open. Which point is at a higher potential: a or b?

Video transcript

Welcome back, everybody. We are tasked with finding what the ideal volt meter reading is going to be when this switch s is open. Well, when that switch s is open, we're really just kind of looking at this loop right here and it's just going to be two batteries in series with some resistors, right? So let's go ahead and apply Kershaw's loop rule to this loop right here. We have that the voltage of this battery battery minus 125 times I which is going to be the voltage of this resistor right here plus volts, which is the voltage of this battery right here -50 times I, which is the voltage of this lower right resistor here is equal to zero, getting like terms together and adding on both sides, we get that 40 is equal to 75 I giving us that our current is equal to 0.53 a. So now what I'm gonna do is I'm going to apply OEMs law between N and M and it states that our voltage of N minus R voltage of M is going to be equal to 10 minus the voltage of this resistor right here, which is 25 times our current of 0.53 plugging this into our calculator. We get negative 3.33 volts which will be the ideal volt meter reading. And since it is negative, let's look at the negative side of the battery which is pointing towards M therefore having the greatest potential. So we have now found the voltage and the fact that M has the greater potential corresponding to our answer choice of C. Thank you all so much for watching. Hope this video helped we will see you all in the next one.