# Simple DC circuits – problems and solutions

1. Based on the figure below, determine the electric current through R_{1}.

__Known :__

Resistor 1 (R_{1}) = 4 Ω

Resistor 2 (R_{2}) = 4 Ω

Resistor 3 (R_{3}) = 8 Ω

Electric voltage (V) = 40 Volt

__Wanted:__ Electric current through R_{1}

__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). After that calculate the electric current using the equation of Ohm’s law :

V = I R or I = V / R

V = voltage, I = electric current, R = the equivalent resistance

__The equivalent resistance :__

Resistor R_{1} and resistor R_{2} are connected in parallel. The equivalent resistance :

1/R_{12} = 1/R_{1} + 1/R_{2 }= 1/4 + 1/4 = 2/4

R_{12 }= 4/2 = 2 Ω

Resistor R_{12} and resistor R_{3} are connected in series. The equivalent resistance :

R = R_{12} + R_{3} = 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.

**Electric voltage **** V**_{ab}** ****and ****V**_{bc}

**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.*

Based on Kirchhoff’s first rule, concluded that if the electric current flows out of the battery are 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.

__Electric voltage ____V___{ab }__:____ __

V_{ab} = I_{ab} R_{ab} = (4)(2) = 8 Volt

__Electric voltage ____V___{bc }__:____ __

V_{bc} = I_{bc} R_{bc} = (4)(8) = 32 Volt

Above circuit are connected in series so that the total electrical voltage is V = V_{ab} + V_{bc} = 8 Volt + 32 Volt = 40 Volt.

**The electric current flows through ****R**_{1}** = 4 Ω**

I_{1} = V_{ab} / R_{1} = 8 Volt / 4 Ohm = 2 A

I_{2} = V_{ab} / R_{2} = 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 R_{1} and the electric current of 2 A flows through the resistor R_{2}. 2 A + 2 A = 4 A.

2. Based on the figure below, determine the electric current that flows through resistor 4-Ω.

__Known :__

Resistor 1 (R_{1}) = 6 Ω

Resistor 2 (R_{2}) = 4 Ω

Resistor 3 (R_{3}) = 1.6 Ω

The electric voltage (V) = 16 Volt

__Wanted:__ The electric current flows through 4 Ω

__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**

__The equivalent resistance :__

Resistor R_{1 }and resistor R_{2 }are connected in parallel. The equivalent resistor :

1/R_{12} = 1/R_{1} + 1/R_{2 }= 1/6 + 1/4 = 2/12 + 3/12 = 5/12

R_{12 }= 12/5 = 2.4 Ω

Resistor R_{12 }and resistor R_{3 }are connected in series. The equivalent resistor :

R = R_{12} + R_{3} = 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 ****V**_{ab }**and ****V**_{bc}

Based on Kirchhoff’s first rule, concluded that if the electric current flows out of the battery are 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 ____V___{ab }__:____ __

V_{ab} = I_{ab} R_{ab} = (4)(2.4) = 9.6 Volt

__The electric voltage ____V___{bc }__:____ __

V_{bc} = I_{bc} R_{bc} = (4)(1.6) = 6.4 Volt

Above circuit are connected in series so that the total electric voltage is V = V_{ab} + V_{bc} = 9.6 Volt + 6.4 Volt = 16 Volt.

**The electric current that flows through ****R**_{2}** = 4 Ω**

I_{1} = V_{ab} / R_{1} = 9.6 Volt / 6 Ohm = 1.6 A

I_{2} = V_{ab} / R_{2} = 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 R_{1} and the electric current 2.4 A flows through the resistor R_{2}. 1.6 A + 2.4 A = 4 A.