Electric circuits – problems and solutions

Electric circuits – problems and solutions

1. R₁, = 6 Ω, R₂ =R₃ = 2 Ω, and voltage = 14 volt, determine the mkondo wa umeme in mzunguko as shown in figure below.

Inajulikana:

Mshambuliaji 1 (R₁) = 6 Ohm

Kipingamizi 2 (R)₂) = 2 Ohm

Kipingamizi 3 (R)₃) = 2 Ohm

voltage (V) = Volti 14

Inayohitajika: Mkondo wa umeme (I)

Suluhisho:

Equivalent resistor (R) :

R₂ na R₃ are connected in parallel. The equivalent resistor :

1 / R₂₃ = 1/R₂ +1/R₃ = 1/2 + 1/2 = 2/2

R₂₃ = 2/2 = 1 Ohm

R₁ na R₂₃ are connected in series. The equivalent resistor :

R=R₁ + R₂₃ = 6 Ω + 1 Ω

R = 7 Ω

Electric current (I) :

I = V / R = 14 / 7 = 2 Ampea

2. Which one of the electric circuits as shown below has the bigger current.

Suluhisho:

The resistance of the resistor is R and the electric voltage is V.

Answer A.

R₁, R₂ na R₃ are connected in series. The equivalent resistor :

R_A =R₁ + R₂ + R₃ = R + R + R = 3R

Electric current (I) :

Answer B.

R₁, R₂ na R₃ are connected in parallel. The equivalent resistor :

1/R = 1/R₁ +1/R₂ +1/R₃ = 1/R + 1/R + 1/R = 3/R

R_B = R/3

Electric current (I) :

Answer C.

R₂ na R₃ are connected in parallel. The equivalent resistor :

1 / R₂₃ = 1/R₂ +1/R₃ = 1/R + 1/R = 2/R

R₂₃ = R/2

R₁ na R₂₃ are connected in series. The equivalent resistor :

R_C =R₁ + R₂₃ = R + R/2 = 2R/2 + R/2 = 3R/2

Electric current (I) :

Answer D.

R₁ na R₂ are connected in parallel. The equivalent resistor :

1 / R₁₂ = 1/R₁ +1/R₂ = 1/R + 1/R = 2/R

R₁₂ = R/2

R₁₂ na R₃ are connected in series. The equivalent resistor :

R_D =R₁₂ + R₃ = R/2 + R = R/2 + 2R/2 = 3R/2

Electric current (I) :

3. R₁ = 4 ohm, R₂ = 6 ohm, R₃ = 2 ohm, and V = 24 volt. What is the electric current in circuit as shown in figure below.

Inajulikana:

Kipingamizi 1 (R)₁) = 4 Ohms

Kipingamizi 2 (R)₂) = 6 Ohms

Kipingamizi 3 (R)₃) = 2 Ohms

Volti (V) = Volti 24

Inayohitajika: Electric current in circuit

Suluhisho:

R₁, R₂ na R₃ are connected in series. The equivalent resistor :

R=R₁ + R₂ + R₃ = 4 + 6 + 2

R = 12 Ohm

Electric current :

I = V / R = 24 / 12 = 2 Ampea

4. Which one of the electric circuits as shown below has the bigger current.

Suluhisho

Electric current in circuit A.

Kipingamizi sawa:

R₁ = 3 Ohm, R₂ = 4 Ohm, R₃ = 4 Ω, V = 12 Volt

R₂ na R₃ are connected in parallel. The equivalent resistor :

1 / R₂₃ = 1/R₂ +1/R₃ = 1/4 + 1/4 = 2/4 = 1/2

R₂₃ = 2/1 = 2 Ohm

R₁ na R₂₃ are connected in series. The equivalent resistor :

R=R₁ + R₂₃ = 3 Ω + 2 Ω = 5 Ω

Electric current (I) :

I = V / R = 12 / 5 = 2.4 Ampea

Electric current in circuit B.

Kipingamizi sawa:

R₁ = 8 Ohm, R₂ = 2 Ohm, R₃ = 2 Ω, V = 36 Volt

R₁, R₂ na R₃ are connected in series. The equivalent resistor :

R=R₁ + R₂ + R₃ = 8 + 2 + 2 = 12 Ω

Electric current (I) :

I = V / R = 36 / 12 = 3 Ampea

Electric current in circuit C.

Kipingamizi sawa:

R₁ = 4 Ohm, R₂ = 4 Ohm, R₃ = 6 Ω, V = 12 Volt

R₂ na R₃ are connected in parallel. The equivalent resistor :

1 / R₂₃ = 1/R₂ +1/R₃ = 1/4 + 1/4 + 1/6 = 3/12 + 3/12 + 2/12 = 8/12

R₂₃ = 12/8 = 1.5 Ohm

Electric current (I) :

I = V / R = 12 / 1.5 = 8 Ampea

Electric current in circuit D.

Kipingamizi sawa:

R₁ = 3 Ohm, R₂ = 3 Ohm, R₃ = 3 Ohm, R₄ = 3 Ohm, R₅ = 6 Ω, V = 24 Volt

R₂, R₃ na R₄ are connected in parallel. The equivalent resistor :

1 / R₂₃₄ = 1/R₂ +1/R₃ +1/R₄ = 1/3 + 1/3 + 1/3 = 3/3

R₂₃₄ = 3/3 = 1 Ohm

R₁, R₂₃₄ na R₅ are connected in series The equivalent resistor :

R=R₁ + R₂₃₄ + R₅ = 3 + 1 + 6 = 9 Ω

Electric current (I) :

I = V / R = 24 / 9 = 2.6 Ampea

5. According to figure as shown below, determine :

A. Total resistance

B. Electric current in circuit

C. Current I₁

D. Current I₂

Inajulikana:

Kipingamizi 1 (R)₁) = 4 Ohm

Kipingamizi 2 (R)₂) = 4 Ohm

Kipingamizi 3 (R)₃) = 2 Ohm

Kipingamizi 4 (R)₄) = 3 Ohm

Electric voltage (V) = 12 Volt

Suluhisho:

A. Total resistance (R)

Kipingamizi R₂ and resistor R₃ are connected in series. The equivalent resistor :

R₂₃ =R₂ + R₃ = 4 Ω + 2 Ω = 6 Ω

Kipingamizi R₂₃ and resistor R₄ are connected in parallel. The equivalent resistor :

1 / R₂₃₄ = 1/R₂₃ +1/R₄ = 1/6 + 1/3 = 1/6 + 2/6 = 3/6

R₂₃₄ = 6/3 = 2 Ohm

Kipingamizi R₁ and resistor R₂₃₄ are connected series. The equivalent resistor :

R=R₁ + R₂₃₄ = 4 Ω + 2 Ω = 6 Ω

The total resistance is 6 Ohm.

B. Electric current in circuit (I)

V = IR

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

Electric current :

I = V / R = 12 Volt / 6 Ohm = 2 Ampere

C. Electric current I₁

Electric current in resistor R₁ = electric current in circuit = 2 Ampere.

D. Current I₂

Kipingamizi R₂₃ and resistor R₄ are connected in parallel. The equivalent resistor R₂₃₄ = 2 Ohm.

Electric current in resistor R₂₃₄ = electric current in resistor R₁ = Ampea 2.

Voltage in resistor R₂₃₄ ni:

V = IR₂₃₄ = (2 A)(2 Ohm) = 4 Volt

Voltage in resistor R₂₃₄ = voltage in resistor R₄ = voltage in resistor R₂₃ = 4 Volt.

The equivalent resistor R₂₃ is 6 Ohm.

Electric current in resistor R₂₃ ni:

I = V / R = 4 Volt / 6 Ohm = 2/3 Ampere

Electric current in resistor R₂₃ = Electric current in resistor R₂ = electric current in resistor R₃ = 2/3 Ampea.

6. R₁ =R₂ = 10 Ω and R₃ =R₄ = 8 Ω. What is the electric current in circuit as shown in figure below ?

Inajulikana:

Kipingamizi R₁ = Resistor R₂ = 10 Ohm

Kipingamizi R₃ = Resistor R₄ = 8 Ohm

Electric voltage (V) = 12 Volt

Inayohitajika: electric current (I)

Suluhisho:

The equivalent resistor

Kipingamizi R₃ and resistor R₄ are connected in parallel, the equivalent resistor :

1 / R₃₄ = 1/R₃ +1/R₄ = 1/8 + 1/8 = 2/8

R₃₄ = 8/2 = 4 Ohm

Kipingamizi R₁, R₂ na R₃₄ are connected in series, the equivalent resistor :

R=R₁ + R₂ + R₃₄ = 10 Ω + 10 Ω + 4 Ω = 24 Ω

Electric current :

I = V / R = 12 Volt / 24 Ohm = 0,5 Volt/Ohm = 0.5 Ampere

7. If the internal resistance of battery ignored, what is the electric current in the circuit shown in figure below.

Inajulikana:

Kipingamizi R₁ = 3 Ohm

Kipingamizi R₂ = 3 Ohm

Kipingamizi R₃ = 6 Ohm

Electric voltage (V) = 6 Volt

Inayohitajika: Mkondo wa umeme (I)

Suluhisho:

Equivalent resistor

Kipingamizi R₁ na R₂ are connected in series. The equivalent resistor :

R₁₂ =R₁ + R₂ = 3 Ohm + 3 Ohm = 6 Ohm

Kipingamizi R₁₂ and resistor 3 are connected in parallel. The equivalent resistor :

1/R = 1/R₁₂ +1/R₃ = 1/6 + 1/6 = 2/6

R = 6/2 = 3 Ohm

Electric current :

I = V / R = 6 / 3 = 2 Ampea

8. What is the total electric current in circuit as shown in figure below.

Inajulikana:

Kipingamizi R₁ = 6 Ohm

Kipingamizi R₂ = 4 Ohm

Electric current (V) = 6 Volt

Internal resistance (r) = 0.6 Ohm

Inayohitajika: Umeme wa sasa

Suluhisho:

Kipingamizi R₁ and resistor R₂ are connected in parallel. The equivalent resistor :

1 / R_P = 1/R₁ +1/R₂ = 1/6 + 1/4 = 4/24 + 6/24 = 10/24

R_P = 24/10 = 2.4 Ohms

Kipingamizi R_P and internal resistance (r) are connected in series. The equivalent resistor :

R=R_P + r = 2.4 Ohm + 0.6 Ohm = 3.0 Ohm

Electric current in circuit :

I = V / R = 6 Volt / 3 Ohm = 2 Ampere

1. What is an electric circuit?
   • Jibu: An electric circuit is a closed path or loop in which electric current can flow continuously. It typically consists of sources of voltage (like batteries), loads (like resistors, LEDs, motors), and conductors to connect them.
2. What distinguishes a series circuit from a parallel circuit?
   • Jibu: In a series circuit, components are connected end-to-end, so there’s a single path for current. In a parallel circuit, components are connected across common points or junctions, providing multiple paths for current.
3. How does Ohm’s Law relate voltage, current, and resistance in a circuit?
   • Jibu: Ohm’s Law states that the current ($I$) flowing through a conductor between two points is directly proportional to the voltage ($V$) across the two points and inversely proportional to the resistance ($R$). It’s represented as I=V/Râ € <.
4. What is the role of a switch in an electric circuit?
   • Jibu: A switch controls the flow of current in a circuit. When closed, it allows current to flow; when open, it interrupts or stops the current flow.
5. Why is a short circuit considered dangerous?
   • Jibu: In a short circuit, the resistance is very low, causing a very high current to flow. This can lead to overheating, fires, or damage to components and should be protected against with fuses or circuit breakers.
6. What is the function of a fuse or a circuit breaker in a circuit?
   • Jibu: Both fuses and circuit breakers are protective devices designed to interrupt a circuit if the current exceeds a predetermined safe level. While fuses “blow” or “melt”, breaking the circuit, circuit breakers “trip”, and can be reset after they interrupt the circuit.
7. How does Kirchhoff’s Current Law (KCL) describe currents at a junction in a circuit?
   • Jibu: Kirchhoff’s Current Law states that the sum of currents entering a junction is equal to the sum of currents leaving that junction. This is essentially a statement of the conservation of electric charge.
8. What is the difference between AC (Alternating Current) and DC (Direct Current)?
   • Jibu: DC refers to the unidirectional flow of electric charge, typically from a battery or a DC power supply. AC, on the other hand, is an electric charge that changes direction periodically, like what’s supplied from the power grid in many countries.
9. What does the term “ground” refer to in electrical circuits?
   • Jibu: “Ground” refers to a reference point in an electrical circuit from which other voltages are measured, or a common return path for electric current, or a direct physical connection to the Earth.
10. Why are capacitors used in electric circuits?

• Jibu: Capacitors store and release electrical energy. They’re used in circuits for various purposes, such as filtering, smoothing voltage fluctuations, coupling and decoupling AC signals, and timing elements in oscillators.