91Ó°ÊÓ

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

Given
We are given the poWer consumed by 6.0 ohms resistance
Solution
(a) Find the current I through the ammeter A. As shown by figure the ammeter is connected between1.0 ohms and 19.ohms, so it reads the current flow in both these resistors. Hence, the target here is to find the current throughthese resistors.
First, let use the given value of the power to find the current flow is 6.0 ohms-The power consumed by the resistance isrelated to the current flows through it in the next form
$$\begin{gathered} \mathrm{P}={\mathrm{I}}^2\mathrm{R} \\ \mathrm{I}=\sqrt{\frac{\mathrm{P}}{\mathrm{R}}} \end{gathered}$$

NowletusplugourvaluesforP69andR69intoequation(1)togetthecurrentthrough6.0ohms

$$\begin{gathered} \mathrm{I}=\sqrt{\frac{\mathrm{P}}{\mathrm{R}}} \\ {\mathrm{I}}_{6\mathrm{Î\copyright }}=\sqrt{\frac{24}{6}} \\ =2\mathrm{A} \end{gathered}$$

Also, let us reduce the two resistors in parallel 20.0 9 and 20.0 9 to make the circuit more simple, where for tworesistorparallel the equivalent resistance of their combination is given by equation 26.3 in the next form

$$\begin{gathered} {\mathrm{R}}_{\mathrm{eq}}=\frac{{\mathrm{R}}_1{\mathrm{R}}_2}{{\mathrm{R}}_1+{\mathrm{R}}_2} \\ =\frac{20×20}{20+20} \\ =10\mathrm{Î\copyright } \end{gathered}$$

Therefore the current through ammeter A is 2A and the volometer reading is 10 ohms

find the current flows through 19.ohms, use the loop rule, where the loop rule is a statement that the electrostatforce is conservativeSuppose around a loop, measuring potential differences across circuit elements as we go anthe algebraic sum of these differences is zero when we return to the starting poin use the right loop in the ?gure and apply equation 26.6 as shown in the figure below where the direction of our tra~is counterclockwise

$$\begin{gathered} \underset{}{∑}\mathrm{V}=0 \\ -25\mathrm{V}+\mathrm{l}6\mathrm{Î\copyright }+{\mathrm{IR}}_{\mathrm{eq}}+\mathrm{l}19+\mathrm{l}1\mathrm{Î\copyright }+{\mathrm{l}}_{6}3=0 \end{gathered}$$

25,Visnegativebecausethedirectionoftravelingisfrompositivetonegativeterminalinthebattery(Seefigure
The terms (I69) (6.0 ohms), I (19.0ohms), I (1.0 ohms) and (I5ohms) (3.0 ohms) are positive because the traveling direction isopposite to the direction of the current (See ?gure 26.8b )-
Note that, the current direction is from the positive terminal of 25 V to up, where a part goes to the right and flow throughReg and a part goes left to 17.0 ohms. Now we can solve the summation for I and we will get
l = 0.233A

Therefore the ammeter A reads 0.233 A

(b) As We discussed in part (a), the current direction is from the positive terminal of 25 V to up, where apart goes to the right and flows through Reg and a part goes left to 17.0 (2, so at this junction point we will apply thejunction rule, where the junction rule is based on conservation of electric charge and the current enters ajunction point isequal to the current shows out from this point, so in our circuit, we could get the current flows in 17.0 ohms, where Current entersthe junction point, while I Re; and I179 flow out from this point
Let us plug our values for I69 and I Reg into equation (2) to get the current flows

$$\begin{gathered} {\mathrm{l}}_6={\mathrm{l}}_{17}+{\mathrm{l}}_{{\mathrm{R}}_{\mathrm{eq}}} \\ {\mathrm{l}}_{17}={\mathrm{l}}_6-{\mathrm{I}}_{\mathrm{R}} \\ {\mathrm{l}}_{17}=2\mathrm{A}-0.233\mathrm{A} \\ =1.767\mathrm{A} \end{gathered}$$

Now it is the time to use the loop rule again but for the left loop to get 8 and apply equation 26.6 where the direction of ourtravel is counterclockwise

25 ,V is positive because the direction of traveling is from negative to positive terminal in the battery, while 8 is negative
because the direction of traveling is from positive to negative terminal in the battery(See ?gure 26.8a )-

$$\begin{gathered} \underset{}{∑}\mathrm{V}=0 \\ -2\mathrm{A}\left(3\right)+25\mathrm{V}-2\mathrm{A}\left(6\right)-1.767\mathrm{A}\left(17\right)-\hat{\mathrm{I}}-7.767\mathrm{A}\left(13\right))=0 \end{gathered}$$

25 ,V is positive because the direction of traveling is from negative to positive terminal in the battery, while is negativebecause the direction of traveling is from positive to negative terminal in the battery(See ?gure 26.8a )-
The terms (2 A) (3.0 ohms), (2 A) (6.0ohms), (1.767 A)(17.0 ohms) and (1.767 A) (13.0 ohsm) are negative because the travelingdirection is the same direction of the current (See ?gure 26.8b ) Now We can solve the summation and we will get
$\hat{\mathrm{l}}=46\mathrm{V}$
The final result is = 46 V but as E: is negative, then the polarity in the figure should be in the opposite direction where thenegative terminal of the battery is next to 17.ohms

therefore the EMF of the battery is -46V and the polarity is in oppostite direction of EMF