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 Sopen. Which point is at a higher potential: aor b? (b) With S closed, find the reading of the voltmeter and the ammeter. Which way (up or down) does the current flow through the switch?

Internal resistance refers to the opposition to the flow of current offered by the cells and batteries themselves resulting in the generation of heat. Internal resistance is measured in Ohms.

(a) When the switch S is open, no current Flows in its path and the ammeter reads zero current while the voltmeter reads thevoltage across path ab. so, our target is to find the voltage across abLet us take the path from a to b Where the direction ofthe current through ab is from up to down as the positive terminal of 25 V is next to aThe potential between a, and b is the sum of the potential drop in the path ab. So let us start from a to b where, in this path,We have resistor 75.ohms and 15 V battery
$$\begin{gathered} {\mathrm{V}}_{\mathrm{ab}}={\mathrm{V}}_{\mathrm{b}}-{\mathrm{V}}_{\mathrm{a}} \\ =-75\mathrm{\Omega l}+15\mathrm{V} \end{gathered}$$

Don'tforgetthattheterms(75.0ohmsI)isnegativebecausethetravelingpaththatWetake,isinthesamedirectionofthecurrentshowsintheresistorand15Vispositivebecausethedirectionoftravelingpathisfromnegativetopositiveterminalof thebattery.
NoW We want to find the current I through 75.ohms. Let us take the left circuit and make a loop in counterclockwise wherethe sum of potential differences across this loop is zero, now apply equation 26.6
$$\begin{gathered} \sum _{}\mathrm{V}=0 \\ -25\mathrm{V}+\mathrm{l}75\mathrm{\Omega }-15\mathrm{V}+\mathrm{l}100\mathrm{\Omega }=0 \end{gathered}$$
25 ,V and 15 ,V are negative because the direction of traveling is from positive to negative terminal in the battery
The terms (I ) (75 ohms) and (I ) (100 ohms) are positive because the traveling direction is in the opposite direction of the current Now we can solve the summation for I and we will getNow we can plug our value for I into equation (1) to get Vab

role="math" localid="1664253906509" $$\begin{gathered} \mathrm{l}=0.288\mathrm{A} \\ {\mathrm{V}}_{\mathrm{ab}}=-75+15\mathrm{V} \\ =-75\mathrm{\Omega }\left(0.228\right)+15\mathrm{V} \\ =-2.10\mathrm{V} \end{gathered}$$

As the potential is negative, therefore, at point a the potential is higher than at point b

(b)WhenSisclosed,thecurrentoutfrom25Vto15VbatteryandS.80thecurrentIinSpathisfromuptodOWnandtheammeterreadsthecurrentthroughS-Inthiscase,wewillapplythejunctionrulewherethecurrentthrough100QequaltsumofthecurrentthroughSandpathab

${\mathrm{l}}_{100}={\mathrm{l}}_{\mathrm{ab}}+{\mathrm{l}}_{\mathrm{s}}$
Our target is to find Is

The voltage drop in 25 V battery is due to 100” so we can use Ohm's law to get the current shows in

$$\begin{gathered} {\mathrm{l}}_{100}=\frac{25\mathrm{V}}{\mathrm{R}} \\ =\frac{25\mathrm{V}}{100} \\ =0.25\mathrm{A} \end{gathered}$$

For Iab apply the loop rule, where We will take the loop betWeen the two paths ab and S in counterclockwise wherethe sum of the potential differences is zero in this loop

$\sum _{}\mathrm{V}=0$

${\mathrm{l}}_{\mathrm{ab}}75\mathrm{\Omega }-15\mathrm{V}=0$

,V is negative because the direction of traveling is from positive to negative terminal in the battery
The terms (Iab) (75 ohms)is positive because the loop direction is in the opposite direction of the current
Now We can solve the summation for Iab and we Will get
${\mathrm{l}}_{\mathrm{ab}}=0.20\mathrm{A}$
Now We can plug our values for Ia], and I100 Q into equation (2) to get the current

$$\begin{gathered} {\mathrm{l}}_{\mathrm{s}}={\mathrm{l}}_{100}-{\mathrm{l}}_{\mathrm{ab}} \\ =0.25\mathrm{A}-0.20\mathrm{A} \\ =0.05\mathrm{A} \end{gathered}$$
Therefore the current is measured bv the ammeter isl 0.05 A
To get the voltage measured by the voltmeter we should calculate the Vab, as in part (a) where the current here is Iab = 0.2so the voltage will be
role="math" localid="1664254227854" $$\begin{gathered} {\mathrm{l}}_{\mathrm{ab}}=0.20 \\ {\mathrm{V}}_{\mathrm{ab}}=-75\mathrm{\Omega l}+15\mathrm{V}=0 \\ =-75\mathrm{\Omega }0.20\mathrm{A}+15\mathrm{V}=0 \end{gathered}$$
Thereforee Ammeter reads 0.05A The voltmeter reads zero voltage