Electrical Questions And Answers

Electrical Question 1

The number of kilowatts in 135 milliwatts is 

1. 1.35 × 10–4 kW
2. 135 × 10–3 kW
3. 0.0135 kW 
4. 0.00135 kW

Answer: A. 1.35 × 10–4 kW

Solution

The number of kilowatts in 135 milliwatts is 

1 kW = 1000 watt 

1 Watt = 10−3

>10^-3 

Kilowatt 

1 milliwatt = 10−3

>10^-3 

watt 

135 milliwatt = 135×10−3

>×10^-3 

watt 

= 135×10−6

>×10^-6 

Kilowatt 

= 1.35×10−4

>×10^-4 

Kilowatt 

Electrical Question 2

An electric heater draws 3.5 A from a 110 V source. The resistance of the heating element is approximately-

1. 385 Ω
2. 38.5 Ω
3. 3.1 Ω
4. 31 Ω

   >Ω

    

Answer: D. 31 Ω

>Ω

 

Solution  

An electric heater is an example of a resistive load. We can apply Ohms law to any passive element. 

According to Ohms law,

V=I.R

>V = I.R

Current drawn from the supply  

I=3.5 A

>I = 3.5 A 

and V=110 V

>V = 110 V

 

R=VI

>R = V/I

R=1103.5=31.428 Ω≡31Ω

>R = 110/3.5 = 31.428 Ω ≡ 31Ω

Electrical Question 3

A certain series circuit consists of a 1/8 W resistor, a 1/4 W resistor, and a 1/2 W resistor. The total resistance is 1200 Ω

>Ω

. If each resistor is operating in the circuit at its maximum power dissipation, the total current flow is 

1. 27 mA
2. 2.7 mA
3. 19 mA
4. 190 mA

Answer: A. 27 mA

Solution 

Total power consumption in the circuit is the sum of all the elements of power consumption. 

Total power consumption =18+14+12=0.875 Watt

>= 1/8+1/4+1/2=0.875 Watt

The total resistance given in the circuit is 1200 Ω

>Ω

Power consumption  

P=V2R

>P = V²/R

0.875×1200=V2

>×1200 = V²

V2=1050 

>V² = 1050 

V=1050=32.4 Volt

>V = √1050=32.4 Volt

Total current= Total voltage in the circuitTotal resistance in the circuit

>Total current= Total voltage in the circuit/Total resistance in the circuit

I=32.41200=0.027 A=27 mA 

>I = 32.4/1200 = 0.027 A = 27 mA 

Electrical Question 4

A series circuit consists of a 4.7 kΩ

>Ω

, a 12 kΩ

>Ω

, and a 2.2 kΩ

>Ω 

resistor. The resistor that has the most voltage drop is 

1. 12 kΩ
2. 2.2 kΩ
3. 4.7 kΩ
4. Impossible to determine from the given information 

Answer: A. 12 kΩ

Solution

A series circuit consists of a 4.7 kΩ

>Ω

, a 12 kΩ

>Ω

, and a 2.2 kΩ resistor. The resistor that has the most voltage drop is dependent on two factors. 

From Ohms law 

V=IR 

The voltage drop is dependent on two factors in a series circuit 

• Current flowing in the element 

• Resistance value 

But in a series circuit, all the elements have the same current, so the maximum voltage drop occurs at, the resistance which is having more value among all. 

∴ 12 KΩ resistance has highest voltage drop. 

Electrical Question 5

The total resistance of a circuit is 680Ω. The percentage of the total voltage appearing across a 47 Ω resistor that makes up part of the total series resistance is 

1. 68%
2. 47%
3. 69%
4. 6.91%

Answer: D. 6.91%

Solution

Total voltage = V 

The total resistance in series combination is R+47 = 680 

R = 680-47 = 633 Ω

From Ohms law 

V=IR

The two-element series circuit is having the same current i.e., total current. 

The total current I=V680

>I=V/680

The voltage drops across 47 Ω is 

V47=V680×47

>V₄₇ = V/680×47

The percentage of the total voltage appearing across a 47 Ω

V47V×100=47680×100=6.91%

>V₄₇/V×100=47/680×100=6.91%

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