1. ከሆነ R1 = 2 Ω፣ R2 = 4 Ω፣ R3 = 6 Ω፣ ይወስኑ የኤሌክትሪክ ኃይል ከታች ባለው ዑደት ውስጥ የሚፈሱ ፍሰቶች።
የሚታወቅ፦
ሞገስ 1 (አር1) = 2 Ω 
ሬዚስተር 2 (አር)2) = 4 Ω
ሬዚስተር 3 (አር)3) = 6 Ω
የemf 1 ምንጭ (ኢ)1) = 9 ቪ
የemf 2 ምንጭ (ኢ)2) = 3 ቪ
የሚፈለግ የኤሌክትሪክ ጅረት (I)
መፍትሔው
This question relates to የኪርቾሆፍ ህግ. How to solve this problem:
የመጀመሪያ ስም, choose the direction of the current. You can decide the opposite current or direction in the clockwise direction.
ሁለተኛ, when the current through the resistor (R) there is a potential decrease so that V = IR signed negative. ሶስተኛ, 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.
– አይአር1 + ኢ1 – አይአር2 – አይአር3 - ኢ2 = 0
– 2 I + 9 – 4 I – 6 I – 3 = 0
– 12 I + 6 = 0
– 12 I = – 6
I = -6 / -12
እኔ = 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.
መፍትሔው
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
እኔ = -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.
መፍትሔው
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
እኔ = 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.
የሚታወቅ፦
ሬዚስተር 1 (አር)1) = 12 Ω
ሬዚስተር 2 (አር)2) = 12 Ω
ሬዚስተር 3 (አር)3) = 3 Ω
ሬዚስተር 4 (አር)4) = 6 Ω
የemf 1 ምንጭ (ኢ)1) = 6 ቮልት
የemf 2 ምንጭ (ኢ)2) = 12 ቮልት
የሚፈለግ፦ The electric current flows in the circuit (I)
መፍትሔው
ተከላካይ 1 (R1) and resistor 2 (R2) are connected in parallel. The equivalent resistor :
1 / አር12 = 1/አር1 + 1/R2 = 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.
– አይአር12 - ኢ1 – አይአር3 - I R4 + ኢ2 = 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.
የሚታወቅ፦
ሬዚስተር 1 (አር)1) = 10 Ω
ሬዚስተር 2 (አር)2) = 6 Ω
ሬዚስተር 3 (አር)3) = 5 Ω
ሬዚስተር 4 (አር)4) = 20 Ω
የemf 1 ምንጭ (ኢ)1) = 8 ቮልት
የemf 2 ምንጭ (ኢ)2) = 12 ቮልት
የሚፈለግ፦ The electric current that flows in circuit
መፍትሔው
ተከላካይ 3 (R3) and resistor 4 (R4) are connected in parallel. The equivalent resistor :
1 / አር34 = 1/አር3 + 1/R4 = 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.
– አይአር1 – አይአር2 - E1 – አይአር34 + ኢ2 = 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.
የሚታወቅ፦
ሬዚስተር 1 (አር)1) = 1 Ω
ሬዚስተር 2 (አር)2) = 6 Ω
ሬዚስተር 3 (አር)3) = 6 Ω
ሬዚስተር 4 (አር)4) = 4 Ω
የemf 1 ምንጭ (ኢ)1) = 12 ቮልት
የemf 2 ምንጭ (ኢ)2) = 6 ቮልት
የሚፈለግ፦ The electric current that flows in circuit
መፍትሔው
ተከላካይ 1 (R1) and resistor 2 (R2) are connected in parallel. The equivalent resistor :
1 / አር12 = 1/አር1 + 1/R2 = 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 - እኔ R12 - ኢ2 - እኔ R4 – አይአር3 = 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
እኔ = 42/76
I = 0.5 A
