Question

$hey.. \: anyone \: of \: you \: plzz \: solve \: 13th \: 18th \: 19th \: 20th.....$





Plzz help me... by solving these questions.. ☺️ ​

Answer 1

Solution :-

13. As per the diagram redrawn in the attachment.

10Ω in Series with 10Ω || 10Ω in Series with 10Ω

→ 20Ω || 20Ω

$= \left( \dfrac{1}{20} + \dfrac{1}{20} \right)^{-1}$

$= \left( \dfrac{2}{20} \right)^{-1}$

$= \left( \dfrac{1}{10} \right)^{-1}$

= 10Ω option (a)

18. As per the diagram redrawn in the attachment.

4Ω in Series with 8Ω || 2Ω in Series with 4Ω

→ 12Ω || 6Ω

$= \left( \dfrac{1}{12} + \dfrac{1}{6} \right)^{-1}$

$= \left( \dfrac{1}{12} + \dfrac{2}{12} \right)^{-1}$

$= \left( \dfrac{3}{12} \right)^{-1}$

$= \left( \dfrac{1}{4} \right)^{-1}$

= 4Ω option (b)

19. According to the loop in the figure

$V_A + 3(i_2) - 6(i_2) - V_B = 0$

$V_A + 3. \dfrac{10}{7} - 6. \dfrac{5}{7} - V_B = 0$

$V_A + \dfrac{30}{7} - \dfrac{30}{7} - V_B = 0$

$V_A - V_B = 0$

Answer = option (d)

20. Consider attachment.

Answer = option (a)