1. Based on the figure below, determine the magetsi kuburikidza R1.
Zvinozivikanwa:
Mudziviriri 1 (R)1) = 4 Ω
Mudziviriri 2 (R)2) = 4 Ω
Resistor 3 (R3) = 8 Ω
Magetsi emagetsi (V) = 40 Volt
Kuda: Electric current through R1
Solution:
The electric current flows from high potential to low potential. The direction of electric current in the circuit above is the same as a clockwise direction.
The electric current that flows out of the battery
First, calculate the equivalent resistance (R). Thereafter, calculate the electric current using the equation of Mutemo waOmm :
V = IR or Ini = V / R
V= voltage, I = magetsi, R = the equivalent resistance
The equivalent resistance :
Mudziviriri R1 uye resistor R2 dzakabatana zvakafanana. The equivalent resistance :
1 / R12 = 1 / R1 +1/R2 = 1/4 + 1/4 = 2/4
R12 = 4/2 = 2 Ω
Mudziviriri R12 uye resistor R3 zvakabatana munhevedzano. The equivalent resistance :
R =R12 +R3 = 2 + 8 = 10 Ω
The electric current that flows out of the battery :
I = V / R = 40 / 10 = 4 A
The electric current that flows out of the battery is 4 Ampere.
Magetsi emagetsi Vab uye Vbc
Kirchhoff‘s first rule states that at any junction point, the sum of all currents entering the junction must equal the sum of all currents leaving the junction.
Maererano ne Mutemo wekutanga waKirchhoff, concluded that if the electric current flows out of the battery is 4 A then the electric current through a-b is equal to 4 Ampere, so also the electric current through b-c is 4 Ampere.
Magetsi emagetsi Vab :
Vab = Iniab Rab = (4)(2) = 8 Mavolts
Magetsi emagetsi Vbc :
Vbc = Inibc Rbc = (4)(8) = 32 Mavolts
The above circuit is connected in series so that the total electrical voltage is V = Vab +Vbc = 8 Mavolts + 32 Mavolts = 40 Mavolts.
The electric current flows through R1 = 4 Ω
I1 =Vab /R1 = 8 Volt / 4 Ohm = 2 A
I2 =Vab /R2 = 8 Volt / 4 Ohm = 2 A
The electric current that flows out of the battery is 4 A. When it arrives at point a the electric current is divided into two, the electric current of 2 Ampere flows through the resistor R1 uye
the electric current of 2 A flows through the resistor R2. 2 A + 2 A = 4 A.
2. Based on the figure below, determine the electric current that flows through resistor 4-Ω.
Zvinozivikanwa:
Mudziviriri 1 (R)1) = 6 Ω
Mudziviriri 2 (R)2) = 4 Ω
Mudziviriri 3 (R)3) = 1.6 Ω
The electric voltage (V) = 16 Volt
Kuda: The electric current flows through 4 Ω
Solution:
The electric current flows from high goneko to low potential. The direction of electric current in the circuit above is the same as a clockwise direction.
The electric current that flows out of the battery
The equivalent resistance :
Mudziviriri R1 uye resistor R2 zvakabatana zvakafanana. Chinodzivirira chakaenzana:
1 / R12 = 1 / R1 +1/R2 = 1/6 + 1/4 = 2/12 + 3/12 = 5/12
R12 = 12/5 = 2.4 Ω
Mudziviriri R12 uye resistor R3 zvakabatana zvakatevedzana. Chinodzivirira chakaenzana:
R =R12 +R3 = 2.4 + 1.6 = 4 Ω
The electric current that flows out of the battery :
I = V / R = 16 / 4 = 4 A
The electric voltage Vab uye Vbc
Based on Kirchhoff’s first rule, concluded that if the electric current flows out of the battery is 4 A then the electric current through a-b is equal to 4 Ampere, so also the electric current through b-c is 4 Ampere.
The electric voltage Vab :
Vab = Iniab Rab = (4)(2.4) = 9.6 Mavolts
The electric voltage Vbc :
Vbc = Inibc Rbc = (4)(1.6) = 6.4 Mavolts
The above circuit is connected in series so that the total electric voltage is V = Vab +Vbc = 9.6 Mavolts + 6.4 Mavolts = 16 Mavolts.
The electric current that flows through R2 = 4 Ω
I1 =Vab /R1 = 9.6 Volt / 6 Ohm = 1.6 A
I2 =Vab /R2 = 9.6 Volt / 4 Ohm = 2.4 A
The electric current that flows out of the battery is 4 A. When it arrives at point a the electric current is divided into two,
the electric current 1.6 A flows through the resistor R1 and the electric current 2.4 A flows through the resistor R2. 1.6 A + 2.4 A = 4 A.
