Question

.. In the circuit shown in Fig. E26.32 both batteries have insignificant internal resistance and the idealized ammeter readin the direction shown. Find the emf E of the battery. Is the polarity shown correct?

Short answer

the EMF of the Battery is 52.35V

Step-by-step solution

About polarity in electric charges

Polarity is defined as the property in a molecule, or compound through which they are either attracted or repelled by an electric chargebecause of an asymmetrical arrangement of electropositive or electronegative atoms around the center of the species

Determine the Current in  the battery

Solution
Find the unknown emf E- The current measured by the ammeter is I = 1.5 A and this is the current flow in
resistance 12 ohms and we will use Kirchhoff's rules in our solution to get E The figure below shows the directions of currentand the traveling paths of loops-
apply the loop rule to get the variables where the loop rule is a statement that the electrostatic force is conservative
Suppose around a loop, measuring potential differences across circuit elements as we go and the algebraic sum oftheese differences is zero when We return to the starting poin use loop 1 and apply equation 26.6 as showin the figure below where the direction of our travel is counterclockwiseto get the current flows in resistance 48 Q which will help us to get the current shows in 15 ohms

$$\begin{gathered} \sum _{}\mathrm{V}=0 \\ -75\mathrm{V}+{\mathrm{l}}_{48\mathrm{\Omega }}+1.5\mathrm{A}12\mathrm{\Omega } \end{gathered}$$

,Visnegativebecausethedirectionoftravelingisfrompositivetonegativeterminalinthebattery(Seefigure26.8a).Theterms(I43ohms)(48ohms)and(1.5A)(12ohms)arepositivebecausethetravelingdirectionisoppositetothedirectionofthe
current Now we can solve the summation for and we will get

${\mathrm{l}}_{48}==1.188\mathrm{A}$
Now l use the junction rule where the junction rule is based on conservation of electric charge and the current enter:ajunction point is equal to the current flOWs out from this point, so in our circuit, We could get the current flow in , whereenters point a, as shown in the figure below while

$$\begin{gathered} \mathrm{l}={\mathrm{l}}_{48}+{\mathrm{l}}_{\mathrm{i}} \\ {\mathrm{l}}_{\mathrm{i}}=\mathrm{l}-{\mathrm{l}}_{48\mathrm{\Omega }} \\ {\mathrm{l}}_{\mathrm{i}}=1.50\mathrm{A}-1.188\mathrm{A} \\ =0.312\mathrm{A} \end{gathered}$$

Determine the EMF of the battery

Again apply the loop rule

Let the outer loop 2 and apply equation 26.6 as shown in the figure below
where the direction of our travel is counterclockwise
:$$\begin{gathered} \sum _{}\mathrm{V}=0 \\ -75\mathrm{V}+{\mathrm{l}}_{\mathrm{i}}15\mathrm{\Omega }+\mathrm{i}+1.5\mathrm{A}12\mathrm{\Omega }=0 \end{gathered}$$

,V is negative because the direction of traveling is from positive to negative terminal in the battery is positive because the direction of traveling is from negative to positive terminal in the batteryThe terms (IE)(15 ohms) and (1.5 A) (12 ohms) are positive because the traveling direction is opposite to the direction of thecurrent in each resistor (See ?gure 26.8b )- Now we can solve the summation for and we will get
The emf is positive, so the polarity in the figure is right

$\mathrm{E}=52.35\mathrm{V}$

Therefore the EMF of the Battery is 52.35V