CircuitKirchhoff's Laws Problems

$$
\begin{array}{l}{\text { a) Use Kirchhoff's laws and Ohm's law to find } \mathrm{I}_{1} \text { in the circuit shown. }} \\ {\text { b) Test the solution for } \mathrm{I}_{1} \text { by verifying that the total power generated equals the total power dissipated. }}\end{array}
$$

خطوات الحل

(1) $$KCL$$ علي أي mode

(2) $$C.L$$ بدون مصدر تيار

(3) $$KVL$$ لو $$
+V\leftarrow(+)
$$

لو $$
-V \leftarrow(-)
$$

* $$KVL$$

(1) لو قابلت تيار في اتجاه المسار $$
V=I R \quad \Leftarrow
$$

(2) لو قابلت التيار عكس اتجاه المسار $$
V=-IR \Leftarrow
$$

(1) $$
\Sigma I=0
$$

$$
I_{1}+I_{2}+6=0 \rightarrow (1)
$$

(2)

(3) $$
\Sigma V=0
$$

$$
-120+I_1*10-I_2*50=0
$$

$$
10 I_{1}-50 I_{2}=120\rightarrow (2)
$$

Solve (1) , (2) $$
I_{1}=\theta 3 A , I_{2}=\theta 3 A
$$

خطوات حساب P

(1) المقاومات

$$
P=I^{2} R \quad(+)
$$

(2) مصادر التيار والجهد

$$
P=V I
$$

$$
P=-6 *12=-72 w
$$

$$
\text { (-) generated }
$$

$$
\text { (+) dissipated }
$$

$$
P=-V I
$$

$$
-\longrightarrow+
$$

$$
3 \longrightarrow
$$

$$
 P_{10}=I^{2} R=(3)^{2} * 10=90 \mathrm{w}
$$

$$
P_{50}=I^{2} R=(3)^{2} * 50=450w
$$

$$
P_{\text {120v }}=V I=120*3=360w
$$

$$
P_{6 A}=-V I=-120 * 6=-900 w
$$

$$
\Sigma P_\text { circuit }=P_{10 \Omega}+P_{50\Omega}+P_{120v}+P_{6A}
$$

$$
=90+450+360-900=0
$$

$$
P_\text { diss }=900w
$$

(1) $$
\Sigma I=0
$$ (any mode)

(2) $$
\Sigma V=0
$$ (any closed coop)

(3) $$
\Sigma P=0
$$ (any circuit)