91Ó°ÊÓ

We want to determine the current 1 in resistor It so we will use the junction rule. At point, the current 1 and enter point a, and the current 6.0 A out from point a so the current 6.0 A is the sum of both current

6A + I + 4A

I = 6A - 4A

= 2A
Therefore the current inresistor is 2A

[b)theresistanceR-TogettheresistanceRwewillapplytheloopruleWheretheloopruleisastatement
that the electrostatic 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

Step I, let us use loop I in the direction counterclockwise (outer blue path)- To make it easier take a

equation 26.5 for loop 1 where the term (2.0 A)R is negative because the
traveling direction is in the direction of the current the emf 28 V is positive because the direction of
traveling is from negative to positive terminal in the batter and the term (6.0 A) (3 Q) is negative
because the traveling direction is in the direction of the current
$$\begin{gathered} \underset{}{∑}\mathrm{V}=0 \\ 28\mathrm{V}-2\mathrm{A}×\mathrm{R}-6\mathrm{A}×3=0 \\ \mathrm{R}=5\mathrm{Î\copyright } \end{gathered}$$

Therefore the resitance in the resistor is 5 ohms

(c) We want to ?nd the unknOWn 8. \Mth the same steps in part (b) but for loop 2 in the direction counterclockwise (red path)

$\begin{array}{l}\underset{}{∑\mathrm{V}}=0\\ 28\mathrm{V}-\hat{\mathrm{i}}+4\mathrm{A}×6\mathrm{A}-2\mathrm{A}×5=0\\ \hat{\mathrm{i}}=42\mathrm{V}\end{array}$

The term is negative because the traveling direction is in the direction of the current (Sethe emf28 V is positive because the direction of traveling is from negative to positive terminal in the battery (See ?gure 26.8a) andthe term (4.0 A) (6 Q) is positive because the traveling direction is in the opposite direction of the current
and the term 8 is negative because the direction of traveling is from positive to negative terminal in the battery (See figure Now We can solve the summation

$∈=42\mathrm{V}$

Therefore the EMF is 42V

(d)Whenthecircuitisbrokenatm,thecurrent?owsinresistorsRand3.09notin6.0Q-Butthecurrentin3.09notequ;
6.0 A and the current in both resistors R and 3.0 Q is I - So let us apply loop equation for loop] in the direction
counterclockwise (Outer blue path)
$$\begin{gathered} \underset{}{∑}\mathrm{V}=0 \\ 28\mathrm{V}-\hat{\mathrm{i}}+4\mathrm{A}×6\mathrm{A}-2\mathrm{A}×5=0 \end{gathered}$$
The term I (3 Q) is negative because the traveling direction is in the direction the emf 28 Vis positive because the direction of traveling is from negative to positive terminal in the battery and theterm is positive because the traveling direction is in the opposite direction of the current Now solve the summation for I and we will get:

i = 3.5 A

positive because the direction of traveling is from negative to positive terminal in the battery is positive because the traveling direction is in the opposite direction of the curret

Therefore the Curret flow through the resistor is 3.5A