Kirchhoff’s rules – problems and solutions

1. What is the terminal voltage of the battery in the circuit below?

Solution

emf = electromotive force = the potential difference between the terminals when no current flow to an external circuit.

The terminal voltage (V) = the potential difference between the terminals when a current flows from the battery.

If no current is drawn from the battery, the terminal voltage equals the emf.

__Known :__

Resistor 1 (R_{1}) = 2 Ω

Resistor 2 (R_{2}) = 4 Ω

Resistor 3 (R_{3}) = 4 Ω

emf 1 (E_{1}) = 20 Volt

emf 2 (E_{2}) = 15 Volt

__Wanted :__ The terminal voltage (V)

__Solution 1 :__

The terminal voltage :

V = E_{1} – E_{2 }= 20 – 15 = 5 Volt

__Solution 2 :__

**Calculate the current flows in the circuit (I) **

* First*, Choose the direction of each current. The direction can be chosen arbitrarily: if the current is actually in the opposite direction, it will come out with a minus sign in the solution.

Then, starting at any point in the circuit, we imagine traveling around a loop, adding emfs and IR terms as we come to them.

* Second*, When we travel through a source in the direction from – to +, the emf is considered to be positive; when we travel from + to -, the emf is considered to be negative.

* Third*, When we travel through a resistor in the same direction as the assumed current, the IR term is negative because the current goes in the direction of decreasing potential. When we travel through a resistor in the direction opposite to the assumed current, the IR term is positive because this represents a rise of potential.

The direction of current is chosen same as the clockwise direction :

E_{1} – I R_{1} – I R_{2} – I R_{3} – E_{2} = 0

20 – I (2) – I (4) – I (4) – 15 = 0

20 – 15 – I (2) – I (4) – I (4) = 0

5 – 10 I = 0

5 = 10 I

I = 5/10

I = 0.5 Ampere

The electric current flowing in the circuit is 0.5 Ampere. Electric current signed positive means the direction of electric current is the same as the clockwise direction.

**Calculate the equivalent resistor (R) :**** **

Resistor 1 (R_{1}), resistor 2 (R_{2}) and resistor 3 (R_{3}) are connected in series. The equivalent resistor :

R = R_{1} + R_{2} + R_{3} = 2 Ω + 4 Ω + 4 Ω = 10 Ω

**The terminal voltage in resistor ****R (V)**

V = I R = (0.5)(10) = 5 Volt

2. In the circuit as shown in figure below, find the power dissipated in the 3-Ω resistor.

Solution :

__Known :__

Resistor 1 (R_{1}) = 2 Ω

Resistor 2 (R_{2}) = 3 Ω

Resistor 3 (R_{3}) = 4 Ω

emf 1 (E_{1}) = 8 Volt

emf 2 (E_{2}) = 10 Volt

__Wanted:__ The power dissipated in the 3-Ω resistor

__Solution :__

The power dissipated in the 3-Ω resistor :

P = V I

*P = **power**, V = **the **voltag**e across **the **3-Ω resistor**, I **= the current passes through the **3-Ω resistor*

**Calculate the electric current ****(I) ****passes through ****the ****3-Ω resistor**

The direction of current is chosen same as the clockwise direction :

E_{1} – I R_{1} – I R_{2} – I R_{3} + E_{2} = 0

8 – I (2) – I (3) – I (4) + 10 = 0

18 – 9 I = 0

18 = 9 I

I = 18 / 9

I = 2 Ampere

The electric current flowing in the circuit is 2 Ampere. Electric current signed positive means the direction of electric current is the same as the clockwise direction.

Circuits are connected in series so that the electric current flowing in the circuit = the electric current passes through the 3-Ω resistor = 2 Ampere.

**Calculate voltage ****(V) ****across the ****3-Ω ****resistor**

V = I R_{2} = (2 A)(3 Ω) = 6 Volt

**The power dissipated in the 3-Ω resistor :**

P = V I = (6 Volt)(2 Ampere) = 12 Volt Ampere = 12 Watt

3. Based on the circuit as shown in the figure below, what is the potential difference between point A and B.

__Known :__

Resistor 1 (R_{1}) = 2 Ω

Resistor 2 (R_{2}) = 3 Ω

Resistor 3 (R_{3}) = 4 Ω

emf 1 (E_{1}) = 8 Volt

emf 2 (E_{2}) = 10 Volt

__Wanted :__ The potential difference (V) between point A and B

__Solution :__

**Calculate the electric current ****(I) ****flowing in the ****3-****Ω ****resistor**** **

The direction of current is chosen same as the clockwise direction :

– E_{1} – I R_{1} – I R_{2} – E_{2} – I R_{3 }= 0

– 8 – I (2) – I (3) – 10 – I (4) = 0

– 18 – 9 I = 0

– 18 = 9 I

I = -18 / 9

I = – 2 Ampere

The electric current flowing in the circuit is 2 Ampere. Electric current signed negative means the direction of electric current counterclockwise direction.

Circuits are connected in series so that the electric current flowing in the circuit = the electric current passes through the 3-Ω resistor = 2 Ampere.

**Calculate the potential difference ****(V) ****across the 3-****Ω ****resistor :**

V = I R_{2} = (2 A)(3 Ω) = 6 Volt

4. Based on the circuit as shown in the figure below, what is the potential difference across the R_{3} resistor.

__Known :__

Resistor 1 (R_{1}) = 2 Ω

Resistor 2 (R_{2}) = 4 Ω

Resistor 3 (R_{3}) = 3 Ω

emf 1 (E_{1}) = 6 Volt

emf 2 (E_{2}) = 9 Volt

__Wanted :__ The potential difference across the R_{3} resistor

__Solution :__

**Calculate the electric current ****(I) ****passes through the ****R**_{3}** ****resistor**

The direction of current is chosen same as the clockwise direction :

E_{1} – I R_{1} – E_{2} – I R_{2} – I R_{3 }= 0

6 – 2I – 9 – 4I – 3I = 0

6 – 9 – 2I – 4I – 3I = 0

-3 – 9I = 0

-3 = 9I

I = -3/9

I = -1/3

The electric current flowing in the circuit is 1/3 Ampere. Electric current signed negative means the direction of electric current counterclockwise direction.

Circuits are connected in series so that the electric current flowing in the circuit = the electric current passes through the R_{3 }resistor = 1/3 Ampere.

**Calculate the potential difference ****(V) ****across the ****R**_{3 }**resistor**** **

V = I R_{3} = (1/3)(3) = 1 Volt

5. The potential electric between point C and D = 4 Volt, find R!

__Known :__

Resistor 1 (R_{1}) = 2 Ω

Resistor 2 (R_{2}) = 2 Ω

Resistor 3 (R_{3}) = R

emf 1 (E_{1}) = 8 Volt

emf 2 (E_{2}) = 4 Volt

The potential difference between C and D (V_{CD}) = 4 Volt

__Wanted :__ R

__Solution :__

V_{CD} = I R

4 = I R

R = 4 / I

…*…………*

**Calculate the electric current ****(I) ****passes through the R resistor **

The direction of current is chosen same as the clockwise direction :

– E_{1} – 2I + E_{2 }– 2I – IR = 0

– 8 – 2I + 4 – 2I – I (4/I) = 0

– 8 – 2I + 4 – 2I – 4 = 0

– 8 + 4 – 4 – 2I – 2I = 0

– 8 – 4I = 0

– 8 = 4I

I = -8 / 4

I = -2 Ampere

The electric current flowing in the circuit is 2 Ampere. Electric current signed negative means the direction of electric current counterclockwise direction.

Circuits are connected in series so that the electric current flowing in the circuit = the electric current passes through the R resistor = 2 Ampere.

**R :**

R = 4 / I = 4 / 2 = 2 Ohm