In the circuit shown in Fig. E26.27, find (a) the current in the 3.00Ω resistor; (b) the unknown emfs E1 and E2; (c) the resistance R. Note that three currents are given.

Electromotive force is defined as the electric potential produced by either electrochemical cell or by changing the magnetic field. EMF is the commonly used acronym for electromotive force.

Solution
(a determine the current in the resistance

so usethejunctionrulewherethejunctionruleis
based on conservation of electric charge. At point below, the current 5.0 A and 3.0 A

enter point a and the
current through 3 ohms out from point a so the current through 3ohms is the sum of both current 5.0 A and 3.0 A

$$\begin{gathered} {\mathrm{I}}_3=5\mathrm{A}+3\mathrm{A} \\ =8\mathrm{A} \end{gathered}$$

Therefore the current in resistor is 8A

(b) for emfs 1 and 2- To get EMF1 apply the loop rule where the loop rule is a statement th:the electrostatic force is conservative. Suppose we go around a loop, measuring potential differences across circuitelements as we go and the algebraic sum of these differences is zero when we return to the starting point
use loop 1 (Closed black path) and apply equation 26.6 where the direction of our travel
is counterclockwise
$$\begin{gathered} \sum _{}\mathrm{V}=0 \\ {\hat{\mathrm{i}}}_1-3\mathrm{A}.4-8\mathrm{A}.3=0 \\ {\hat{\mathrm{l}}}_1=36\mathrm{V} \end{gathered}$$

Therefore the EMF 1 is 36V

To get EMF2, use loop (2) (Closed red path) and apply equation 26.6 as shoWn in the figure below where the direction ofour travel is counterclockwise

$$\begin{gathered} \sum _{}\mathrm{V}=0 \\ -\hat{\mathrm{i}}+8\mathrm{A}.3+5\mathrm{A}.6=0 \\ {\hat{\mathrm{i}}}_2=54\mathrm{V} \end{gathered}$$

Therefore the EMF 2 is 54V

EMF1 is negative because the direction of traveling is from positive to negative terminal in the battery (See ?gure 26.8a ).The terms (5.0 A) (6 ohms) and (8.0 A) (3 ohms) are positive because the traveling direction is in the opposite direction of the
current (See ?gure 26.8b )-

Now We can solve the summation for EMF2 and we will get

EMF=54V

Therefore the EMF 2 is 54V

(c)todeterminethevalueofR-Againusetheloopruleandwewillusetheouterloop(3)(Closedblue
path) as shOWn in the ?gure below. Apply equation 26.6 for this loop to get the value of

$$\begin{gathered} \sum _{}\mathrm{V}=0 \\ -3\mathrm{A}.4+5\mathrm{A}.6-2\mathrm{AR}=0 \\ \mathrm{R}=9\mathrm{\Omega } \end{gathered}$$
The terms (3.0 A) (4 ohms) and (2.0 A)R are negative because the traveling direction is the same direction of the current and the term (5.0 A) (6 ohms) is positive because the traveling direction is in the opposite direction of the current- Now we can solve the summation for R and we will get

$\mathrm{R}=9\mathrm{\Omega }$

Therefore the resitance is$\mathrm{R}=9\mathrm{\Omega }$