કિર્ચહોફ કાયદો - સમસ્યાઓ અને ઉકેલો

૧. જો આર₁ = 2 Ω, આર₂ = 4 Ω, આર₃ = 6 Ω, નક્કી કરો વીજ પ્રવાહ નીચેના સર્કિટમાં વહે છે.

જાણીતા:

રેઝિસ્ટર 1 (આર₁) = 2 Ω

રેઝિસ્ટર 2 (R₂) = 4 Ω

રેઝિસ્ટર 3 (R₃) = 6 Ω

emf 1 (E) નો સ્ત્રોત₁) = 9 વી

emf 2 (E) નો સ્ત્રોત₂) = 3 વી

ઇચ્છિત: Electric current (I)

ઉકેલો:

This question relates to કિર્ચહોફનો કાયદો. How to solve this problem:

પ્રથમ, choose the direction of the current. You can decide the opposite current or direction in the clockwise direction.

બીજું, when the current through the resistor (R) there is a potential decrease so that V = IR signed negative. થર્ડ, if the current moves from low to high voltage (- to +) then the source of emf (E) signed positive because of the charging of energy at the emf source. If the current moves from high to low voltage (+ to -) then the source of emf (E) signed negative because of the emptying of energy at the emf source.

In this solution, the direction of the current is the same as the direction of clockwise rotation.

- આઈઆર₁ + ઇ₁ - આઈઆર₂ - આઈઆર₃ - ઇ₂ = 0

– 2 I + 9 – 4 I – 6 I – 3 = 0

– ૫૦૦ I + ૧૨ = ૦

– ૫૦૦ I = – ૧૨

I = -6 / -12

હું = 0.5

The electric current flows in the circuit are 0.5 A. The electric current signed positive means that the direction of the electric current is the same as the direction of clockwise rotation. If the electric current is negative, then the electric current is opposite the clockwise direction.

2. Determine the electric current that flows in the circuit as shown in the figure below.

ઉકેલો:

In this solution, the direction of the current is the same as the direction of clockwise rotation.

-20 – 5I -5I – 12 – 10I = 0

-૩ – ૯I = ૦

-૩ = ૯આઈ

હું = -૧૮ / ૯

I = -1.6 A

Because the electric current is negative, the direction of the electric current is actually opposite to the clockwise direction. The direction of electric current is not the same as estimation.

3. Determine the electric current that flows in the circuit as shown in the figure below.

ઉકેલો:

In this solution, the direction of current is the same as the direction of clockwise rotation.

– I – 6I + 12 – 2I + 12 = 0

-9I + 24 = 0

-9I = -24

હું = ૧,૦૦૦,૦૦૦ / ૮૩૧.૩૬

I = 8 / 3 A

4. An electric circuit consists of four resistors, R₁ = 12 Ohm, R₂ = 12 Ohm, R₃ = 3 Ohm and R₄ = 6 Ohm, are connected with source of emf E₁ = 6 Volt, E₂ = 12 Volt. Determine the electric current flows in the circuit as shown in figure below.

જાણીતા:

રેઝિસ્ટર 1 (R₁) = 12 Ω

રેઝિસ્ટર 2 (R₂) = 12 Ω

રેઝિસ્ટર 3 (R₃) = 3 Ω

રેઝિસ્ટર 4 (R₄) = 6 Ω

emf 1 (E) નો સ્ત્રોત₁) = 6 વોલ્ટ

emf 2 (E) નો સ્ત્રોત₂) = 12 વોલ્ટ

જોઈતું હતું: The electric current flows in the circuit (I)

ઉકેલો:

રેઝિસ્ટર ૧ (R₁) and resistor 2 (R₂) are connected in parallel. The equivalent resistor :

1 / આર₁₂ = 1/R₁ +1/આર₂ = ૧/૬ + ૧/૬ = ૨/૬

R₁₂ = ૬/૨ = ૩ Ω

In this solution, the direction of current is the same as the direction of clockwise rotation.

- આઈઆર₁₂ - ઇ₁ - આઈઆર₃ - I R₄ + ઇ₂ = 0

– 6 I – 6 – 3I – 6I + 12 = 0

– 6I – 3I – 6I = 6 -12

– 15I = – 6

I = -6/-15

I = 2/5 A

5. Determine the electric current that flows in circuit as shown in figure below.

જાણીતા:

રેઝિસ્ટર 1 (R₁) = 10 Ω

રેઝિસ્ટર 2 (R₂) = 6 Ω

રેઝિસ્ટર 3 (R₃) = 5 Ω

રેઝિસ્ટર 4 (R₄) = 20 Ω

emf 1 (E) નો સ્ત્રોત₁) = 8 વોલ્ટ

emf 2 (E) નો સ્ત્રોત₂) = 12 વોલ્ટ

જોઈતું હતું: The electric current that flows in circuit

ઉકેલો:

રેઝિસ્ટર ૧ (R₃) and resistor 4 (R₄) are connected in parallel. The equivalent resistor :

1 / આર₃₄ = 1/R₃ +1/આર₄ = ૧/૪૦૦ + ૧/૨૦૦ = ૧/૪૦૦ + ૨/૪૦૦ = ૩/૪૦૦

R₃₄ = ૬/૨ = ૩ Ω

In this solution, the direction of current is the same as the direction of clockwise rotation.

- આઈઆર₁ - આઈઆર₂ - E₁ - આઈઆર₃₄ + ઇ₂ = 0

– 10I – 6I - 8 - 4I + 12 = 0

– 10I – 6I - 4I = 8 - 12

- 20I = - 4

I = -4/-20

I = 1/5 A

I = 0.2 A

6. Determine the electric current that flows in circuit as shown in figure below.

જાણીતા:

રેઝિસ્ટર 1 (R₁) = 1 Ω

રેઝિસ્ટર 2 (R₂) = 6 Ω

રેઝિસ્ટર 3 (R₃) = 6 Ω

રેઝિસ્ટર 4 (R₄) = 4 Ω

emf 1 (E) નો સ્ત્રોત₁) = 12 વોલ્ટ

emf 2 (E) નો સ્ત્રોત₂) = 6 વોલ્ટ

જોઈતું હતું: The electric current that flows in circuit

ઉકેલો:

રેઝિસ્ટર ૧ (R₁) and resistor 2 (R₂) are connected in parallel. The equivalent resistor :

1 / આર₁₂ = 1/R₁ +1/આર₂ = ૧/૪૦૦ + ૧/૨૦૦ = ૧/૪૦૦ + ૨/૪૦૦ = ૩/૪૦૦

R₁₂ = ૧૦/૩ Ω

The direction of current is the same as the direction of clockwise rotation.

E₁ - હું R₁₂ - ઇ₂ - હું R₄ - આઈઆર₃ = 0

12 – (6/7)I – 6 – 4I – 6I = 0

12 – 6 – (6/7)I – 4I – 6I = 0

6 – (6/7)I – 10I = 0

6 = (6/7)I + 10I

6 = (6/7)I + (70/7)I

6 = (76/7)I

(6)(7) = 76I

42 = 76I

હું = ૫/૧૦

I = 0.5 A