Question

Find the emfs E1 and E2 in the circuit of Fig. E26.26, and find the potential difference of point brelative to point a.

Short answer

the current crossing the EMF is 1A

EMF in first is 18V

the EMF 2 is 7V

the potential difference is 13V

Step-by-step solution

About Emf And potential difference

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.

The potential difference is the difference in potential between two points that represents the work involved or the energy released in the transfer of a unit quantity of electricity from one point to the other

Determine the Current  in the circuit

ApplytheloopruleWheretheloopruleisastatementthattheelectrostati
force is conservative Suppose We go around a loop, measuring potential differences across circuit elements as We go and
the algebraic sum of these differences is zero When We return to the starting point
the current between the center path ab so We will use the junction rule where junction rule is base:on conservation of electric charge.point a so the current is the sum of both current
$$\begin{gathered} {\mathrm{l}}_{\mathrm{ab}}=1.0\mathrm{A} \\ 2\mathrm{A} \\ 2\mathrm{A}={\mathrm{l}}_{\mathrm{ab}}+1\mathrm{A} \\ {\mathrm{l}}_{\mathrm{ab}}=2\mathrm{A}-1\mathrm{A} \\ \mathrm{I}=1.\mathrm{A} \end{gathered}$$
Therfore the current crossing the EMF is 1A

Determine the EMF

loop 2 in the direction counterclockwise (red closed path ). To make it easier take a look
854- Let us apply equation

$$\begin{gathered} \sum _{}\mathrm{V}=0 \\ 20\mathrm{V}-{\hat{\mathrm{i}}}_1-\mathrm{A}1\mathrm{\Omega }+{\mathrm{l}}_{\mathrm{ab}}4\mathrm{\Omega }-1\mathrm{A}6\mathrm{\Omega }=0 \end{gathered}$$

The emf 20 V is positive because the direction of traveling is from negative to positive terminal

rminal in the battery is negative because the direction of traveling is from positive to negative terminal in the battery
The terms 1.0 A(1 9) and (1.0 A) (6ohms) are negative because the traveling direction is in the direction of the current
The terms Iab(4 ohms) and (Iab) (1ohms) are positive because the traveling direction is in the opposite direction of the current
Now solve the summation for E1 and we will get

${\in }_0=18\mathrm{V}$

Therefore EMF in first is 18V

:Determine the EMF in second

useloopIinthedirectioncounterclockwise(blackclosedpath)andapplyequation26-
$$\begin{gathered} \sum _{}\mathrm{V}=0 \\ -{\hat{\mathrm{i}}}_2-2\mathrm{A}2\mathrm{\Omega }+{\hat{\mathrm{i}}}_1-{\mathrm{l}}_{\mathrm{ab}}1\mathrm{\Omega }-{\mathrm{l}}_{\mathrm{ab}}4\mathrm{\Omega }=0-{\hat{\mathrm{i}}}_2=7\mathrm{V} \end{gathered}$$
ihe emf one is positive because the direction of traveling is from negative to positive terminal in the battery is negative because the direction of traveling is from positive to negative terminal in the battery -ihe terms (2.0 A) (2 ohms), (Iab) (1 ohms) and (Iab) (4 ohms) are negative because the traveling direction is in the direction of the
:urrent - Now solve the summation for EMF 2 and we get

${\hat{\mathrm{i}}}_2=7\mathrm{V}$

Therefore the EMF 2 is 7V

:Determine the potential difference between b and a

To get the voltage between points a. and b. Let us take the center path between a, and b Where our travel Will begin from a to b and We Will go through EMF1, 1 ohms and 4 ohms- And the direction of the traveling is opposite to the current ?ow
n 1 ohms and 4 so they will be positive in our calculation
Also, the direction of the traveling is from positive to negative terminal in the battery so it will be negative The voltage Vab is the summation of these three elements in the

$$\begin{gathered} {\mathrm{V}}_{\mathrm{a}}={\mathrm{l}}_{\mathrm{ab}}4+{\mathrm{l}}_{\mathrm{ab}}1-{\hat{\mathrm{i}}}_1 \\ =1\mathrm{A}4+11-18\mathrm{V} \\ =-13\mathrm{V} \end{gathered}$$
The voltage result is negative which indicates to us that the potential at b is lOWer than at a and the potential across ab is

Therefore the potential difference is 13V