Find current in each resistor I1 and I2

Tutaweka mikondo mitatu ya kitanzi, $i_1$, $i_2$, na $i_3$, zote zikizunguka kwa mwelekeo wa saa (clockwise) kama inavyoonyeshwa kwenye mchoro.

• KVL kwa Kitanzi cha Kushoto (Loop 1): Kuanzia kona ya chini kushoto na kusonga kwa mwelekeo wa saa:

  $$-10\text{V} + (3\Omega)i_1 + (5\Omega)(i_1 - i_2) = 0$$ $$3i_1 + 5i_1 - 5i_2 = 10$$ $$8i_1 - 5i_2 = 10 \quad \text{(Equation 1)}$$

• KVL kwa Kitanzi cha Kati (Loop 2): Kuanzia katikati chini na kusonga kwa mwelekeo wa saa:

  $$-(5\Omega)(i_1 - i_2) - 15\text{V} + (9\Omega)(i_2 - i_3) = 0$$ $$-5i_1 + 5i_2 - 15 + 9i_2 - 9i_3 = 0$$ $$-5i_1 + 14i_2 - 9i_3 = 15 \quad \text{(Equation 2)}$$

• KVL kwa Kitanzi cha Kulia (Loop 3): Kuanzia kona ya chini kulia na kusonga kwa mwelekeo wa saa:

  $$-(9\Omega)(i_2 - i_3) + (4\Omega)i_3 + 25\text{V} = 0$$ $$-9i_2 + 9i_3 + 4i_3 = -25$$ $$-9i_2 + 13i_3 = -25 \quad \text{(Equation 3)}$$

Tuna mfumo wa milinganyo mitatu:

1. $8i_1 - 5i_2 = 10$
2. $-5i_1 + 14i_2 - 9i_3 = 15$
3. $-9i_2 + 13i_3 = -25$

Kutoka Mlinganyo (1), tunaweza kutoa $i_1$:

$$i_1 = \frac{10 + 5i_2}{8}$$

Kutoka Mlinganyo (3), tunaweza kutoa $i_3$:

$$i_3 = \frac{-25 + 9i_2}{13}$$

Badilisha $i_1$ na $i_3$ kwenye Mlinganyo (2):

$$-5\left(\frac{10 + 5i_2}{8}\right) + 14i_2 - 9\left(\frac{-25 + 9i_2}{13}\right) = 15$$

Ili kuondoa madhehebu, zidisha mlinganyo mzima kwa $8 \times 13 = 104$:

$$-5 \times 13 (10 + 5i_2) + 14 \times 104 i_2 - 9 \times 8 (-25 + 9i_2) = 15 \times 104$$ $$-65(10 + 5i_2) + 1456i_2 - 72(-25 + 9i_2) = 1560$$ $$-650 - 325i_2 + 1456i_2 + 1800 - 648i_2 = 1560$$

Kusanya vigezo vya $i_2$ na namba za kawaida:

$$(1456 - 325 - 648)i_2 + (1800 - 650) = 1560$$ $$483i_2 + 1150 = 1560$$ $$483i_2 = 1560 - 1150$$ $$483i_2 = 410$$ $$i_2 = \frac{410}{483} \approx 0.84886 \text{ A}$$

Sasa tafuta $i_1$:

$$i_1 = \frac{10 + 5i_2}{8} = \frac{10 + 5(410/483)}{8} = \frac{10 + 2050/483}{8} = \frac{(4830 + 2050)/483}{8} = \frac{6880}{3864} \approx 1.7805 \text{ A}$$

Sasa tafuta $i_3$:

$$i_3 = \frac{-25 + 9i_2}{13} = \frac{-25 + 9(410/483)}{13} = \frac{-25 + 3690/483}{13} = \frac{(-12075 + 3690)/483}{13} = \frac{-8385}{6279} \approx -1.3354 \text{ A}$$

Mkondo $I_2$ umeonyeshwa ukielekea kushoto kupitia kipingamizi cha $5\Omega$. Kwa kutumia mikondo ya kitanzi, mkondo unaoelekea kushoto ni $i_2 - i_1$.

$$I_2 = i_2 - i_1 = \frac{410}{483} - \frac{6880}{3864} = \frac{3280}{3864} - \frac{6880}{3864} = \frac{-3600}{3864}$$ $$I_2 \approx -0.9317 \text{ A}$$

Ishara hasi inaonyesha kuwa mkondo halisi unapita kinyume na mwelekeo wa mshale wa $I_2$ kwenye mchoro (yaani, kuelekea kulia).

Mkondo $I_1$ umeonyeshwa ukielekea chini kupitia kipingamizi cha $9\Omega$. Kwa kutumia mikondo ya kitanzi, mkondo unaoelekea chini ni $i_2 - i_3$.

$$I_1 = i_2 - i_3 = \frac{410}{483} - \left(\frac{-8385}{6279}\right) = \frac{410}{483} + \frac{8385}{6279}$$ $$I_1 = \frac{5330}{6279} + \frac{8385}{6279} = \frac{13715}{6279}$$ $$I_1 \approx 2.1842 \text{ A}$$

Ishara chanya inaonyesha kuwa mkondo halisi unapita katika mwelekeo sawa na mshale wa $I_1$ kwenye mchoro.

Mikondo katika kila kipingamizi ni:
$I_1 = \boxed{2.18 \text{ A}}$
$I_2 = \boxed{-0.93 \text{ A}}$