1. ਜੇਕਰ ਆਰ1 = 2 Ω, ਆਰ2 = 4 Ω, ਆਰ3 = 6 Ω, ਨਿਰਧਾਰਤ ਕਰੋ ਬਿਜਲੀ ਦਾ ਕਰੰਟ ਹੇਠਾਂ ਦਿੱਤੇ ਸਰਕਟ ਵਿੱਚ ਵਹਿੰਦਾ ਹੈ।
ਜਾਣਿਆ ਜਾਂਦਾ ਹੈ:
ਰੋਜ਼ਰ 1 (ਆਰ1) = 2 Ω 
ਰੋਧਕ 2 (R2) = 4 Ω
ਰੋਧਕ 3 (R3) = 6 Ω
emf 1 (E) ਦਾ ਸਰੋਤ1) = 9 ਵੀ
emf 2 (E) ਦਾ ਸਰੋਤ2) = 3 ਵੀ
ਲੋੜੀਂਦਾ: ਬਿਜਲੀ ਦਾ ਕਰੰਟ (I)
ਹੱਲ:
This question relates to Kirchhoff ਦਾ ਕਾਨੂੰਨ. How to solve this problem:
ਪਹਿਲੀ, choose the direction of the current. You can decide the opposite current or direction in the clockwise direction.
ਦੂਜਾ, when the current through the resistor (R) there is a potential decrease so that V = IR signed negative. ਤੀਜਾ, 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.
- ਆਈ.ਆਰ.1 + ਈ1 - ਆਈ.ਆਰ.2 - ਆਈ.ਆਰ.3 - ਈ2 = 0
– 2 I + 9 – 4 I – 6 I – 3 = 0
– 12 I + 6 = 0
– 12 I = – 6
I = -6 / -12
ਮੈਂ = 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.
ਹੱਲ:
In this solution, the direction of the current is the same as the direction of clockwise rotation.
-20 – 5I -5I – 12 – 10I = 0
-32 – 20I = 0
-32 = 20I
ਮੈਂ = -32 / 20
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.
ਹੱਲ:
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
ਮੈਂ = 24/9
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.
ਜਾਣਿਆ ਜਾਂਦਾ ਹੈ:
ਰੋਧਕ 1 (R1) = 12 Ω
ਰੋਧਕ 2 (R2) = 12 Ω
ਰੋਧਕ 3 (R3) = 3 Ω
ਰੋਧਕ 4 (R4) = 6 Ω
emf 1 (E) ਦਾ ਸਰੋਤ1) = 6 ਵੋਲਟ
emf 2 (E) ਦਾ ਸਰੋਤ2) = 12 ਵੋਲਟ
ਲੋੜੀਂਦਾ: The electric current flows in the circuit (I)
ਹੱਲ:
ਰੋਧਕ 1 (R1) and resistor 2 (R2) are connected in parallel. The equivalent resistor :
1 / R12 = 1/ਆਰ1 + 1/ਆਰ2 = 1/12 + 1/12 = 2/12
R12 = 12/2 = 6 Ω
In this solution, the direction of current is the same as the direction of clockwise rotation.
- ਆਈ.ਆਰ.12 - ਈ1 - ਆਈ.ਆਰ.3 - ਆਈ.ਆਰ4 + ਈ2 = 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.
ਜਾਣਿਆ ਜਾਂਦਾ ਹੈ:
ਰੋਧਕ 1 (R1) = 10 Ω
ਰੋਧਕ 2 (R2) = 6 Ω
ਰੋਧਕ 3 (R3) = 5 Ω
ਰੋਧਕ 4 (R4) = 20 Ω
emf 1 (E) ਦਾ ਸਰੋਤ1) = 8 ਵੋਲਟ
emf 2 (E) ਦਾ ਸਰੋਤ2) = 12 ਵੋਲਟ
ਲੋੜੀਂਦਾ: The electric current that flows in circuit
ਹੱਲ:
ਰੋਧਕ 3 (R3) and resistor 4 (R4) are connected in parallel. The equivalent resistor :
1 / R34 = 1/ਆਰ3 + 1/ਆਰ4 = 1/5 + 1/20 = 4/20 + 1/20 = 5/20
R34 = 20/5 = 4 Ω
In this solution, the direction of current is the same as the direction of clockwise rotation.
- ਆਈ.ਆਰ.1 - ਆਈ.ਆਰ.2 - E1 - ਆਈ.ਆਰ.34 + ਈ2 = 0
– 10I – 6I - 8 - 4I + 12 = 0
– 10I – 6I - 4I = 8 - 12
- 20I = – 4
I = -4/-20
I = 1/5 A
ਆਈ = 0.2 ਏ
6. Determine the electric current that flows in circuit as shown in figure below.
ਜਾਣਿਆ ਜਾਂਦਾ ਹੈ:
ਰੋਧਕ 1 (R1) = 1 Ω
ਰੋਧਕ 2 (R2) = 6 Ω
ਰੋਧਕ 3 (R3) = 6 Ω
ਰੋਧਕ 4 (R4) = 4 Ω
emf 1 (E) ਦਾ ਸਰੋਤ1) = 12 ਵੋਲਟ
emf 2 (E) ਦਾ ਸਰੋਤ2) = 6 ਵੋਲਟ
ਲੋੜੀਂਦਾ: The electric current that flows in circuit
ਹੱਲ:
ਰੋਧਕ 1 (R1) and resistor 2 (R2) are connected in parallel. The equivalent resistor :
1 / R12 = 1/ਆਰ1 + 1/ਆਰ2 = 1/1 + 1/6 = 6/6 + 1/6 = 7/6
R12 = 6/7 Ω
The direction of current is the same as the direction of clockwise rotation.
E1 - ਮੈਂ R12 - ਈ2 - ਮੈਂ R4 - ਆਈ.ਆਰ.3 = 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
ਮੈਂ = 42/76
ਆਈ = 0.5 ਏ
