Circuit theory

Unit of a charge is

a) ampere

b) volt

c) coloumb

d) watt

Correct Answer: Option C

Solution:

Unit of charge is coloumb

Circuit theory

Unit of a charge is

a) ampere

b) volt

c) coloumb

d) watt

Correct Answer: Option C

Solution:

Unit of charge is coloumb

Circuit Theory

Which of the following amount of electrons is equivalent to $-3.941$ C of charge?

a) $1.628×10^{20}$

b) $1.24×10^{18}$

c) $6.482×10^{17}$

d) $2.46×10^{19}$

Correct Answer: Option D

Solution:

As we know, $6.24×10^{18}$ electrons consitute $-1$C of charge. Hence $-3.941$C of charge has $3.941×6.24×10^{18}$ no of electrons. $$=2.457×10^{19}$$

Circuit Theory

Which of the following amount of electrons is equivalent to $-3.941$ C of charge?

a) $1.628×10^{20}$

b) $1.24×10^{18}$

c) $6.482×10^{17}$

d) $2.46×10^{19}$

Correct Answer: Option D

Solution:

As we know, $6.24×10^{18}$ electrons consitute $-1$C of charge. Hence $-3.941$C of charge has $3.941×6.24×10^{18}$ no of electrons. $$=2.457×10^{19}$$

Circuit Theory

Which of the following equation is correct?

a) $I=Q×t$

b) $I=\frac{Q}{t}$

c) $Q=\frac{I}{t}$

d) None of these

Correct Answer: Option B

Solution:

Current flow is nothing but rate of flow of charge $$I=\frac{Q}{t}$$

Circuit Theory

Which of the following equation is correct?

a) $I=Q×t$

b) $I=\frac{Q}{t}$

c) $Q=\frac{I}{t}$

d) None of these

Correct Answer: Option B

Solution:

Current flow is nothing but rate of flow of charge $$I=\frac{Q}{t}$$

Circuit Theory

Unit of current is

a) ampere

b) coloumbs/sec

c) Both a and b

d) None of these

Correct Answer: Option C

Solution:

Unit of current is ampere (A). Also $$I=\frac{Q}{t}$$ Hence unit of current is also coloumbs/sec

Circuit Theory

Unit of current is

a) ampere

b) coloumbs/sec

c) Both a and b

d) None of these

Correct Answer: Option C

Solution:

Unit of current is ampere (A). Also
$$I=\frac{Q}{t}$$

Hence unit of current is also coloumbs/sec

Circuit Theory

If $5$A of electric current flows for a period of $3$ minutes, what will be the amount of charge transferred?

a) $600$C

b) $60$C

c) $100$C

d) $900$C

Correct Answer: Option D

Solution:

As we know $$I=\frac{Q}{t}$$I in amperes, Q in coloumbs, t in seconds.$$Q=I×t$$ $$Q=5×3×60$$ $$Q=900C$$

Circuit Theory

If $5$A of electric current flows for a period of $3$ minutes, what will be the amount of charge transferred?

a) $600$C

b) $60$C

c) $100$C

d) $900$C

Correct Answer: Option D

Solution:

As we know $$I=\frac{Q}{t}$$I in amperes, Q in coloumbs, t in seconds.$$Q=I×t$$ $$Q=5×3×60$$ $$Q=900C$$

Circuit Theory

Unit of voltage is/are

a) volt

b) joules/coloumb

c) Nm/C

d) All the above

Correct Answer: Option D

Solution:

Voltage is nothing but how much Work-done is needed in moving a unit charge from infinity to a point against the influence of electric field. 

Voltage = Work-done per unit charge$$V (t)=\frac {dW}{dq}$$ 

Unit:

• volt (V) 
• Joules/Coloumb (since work-done per unit charge) 
• Also, Work done = Force × distance, hence unit of voltage can be Nm/C (Newton - metre / Coloumb)

Circuit Theory

Unit of voltage is/are

a) volt

b) joules/coloumb

c) Nm/C

d) All the above

Correct Answer: Option D

Solution:

Voltage is nothing but how much Work-done is needed in moving a unit charge from infinity to a point against the influence of electric field. 

Voltage = Work-done per unit charge$$V (t)=\frac {dW}{dq}$$ 

Unit:

• volt (V) 
• Joules/Coloumb (since work-done per unit charge) 
• Also, Work done = Force × distance, hence unit of voltage can be Nm/C (Newton - metre / Coloumb)
  
  

Circuit Theory

 Unit of power is/are

a) watts

b) joules/sec

c) Both a and b

d) None of the above

Correct Answer: Option C

Solution:

Power

• Rate at which the work is done
• Rate of expending or consuming energy. $$p (t)=\frac{dW}{dt}$$
• Unit:
• Watt
• Joules/second (since Work-done per unit time)
• Also, Work done = Force × distance, hence unit of power can be Nm/s

Circuit Theory

 Unit of power is/are

a) watts

b) joules/sec

c) Both a and b

d) None of the above

Correct Answer: Option C

Solution:

Power

• Rate at which the work is done
• Rate of expending or consuming energy. $$p (t)=\frac{dW}{dt}$$
• Unit:
• Watt
• Joules/second (since Work-done per unit time)
• Also, Work done = Force × distance, hence unit of power can be Nm/s

Circuit Theory

 The SI unit of power is

a) watt

b) joule

c) newton

d) hertz

Correct Answer: Option A

Circuit Theory

 The SI unit of power is

a) watt

b) joule

c) newton

d) hertz

Correct Answer: Option A

Circuit Theory

 In MKS units, watt is

a) joules/sec 

b) joules-sec

c) joules/sec²

d) none of these

Correct Answer: Option A

Circuit Theory

 In MKS units, watt is

a) joules/sec 

b) joules-sec

c) joules/sec²

d) none of these

Correct Answer: Option A

Circuit Theory

One horsepower (HP) is equal to

a) $7.46$ watts

b) $74.6$ watts

c) $746$ watts

d) $0.746$ watts

Correct Answer: Option C

Solution:

One horsepower is equal to $746$ watts

Circuit Theory

One horsepower (HP) is equal to

a) $7.46$ watts

b) $74.6$ watts

c) $746$ watts

d) $0.746$ watts

Correct Answer: Option C

Solution:

One horsepower is equal to $746$ watts

Circuit Theory

Match the following

A: Power        $1$: Work-done in moving an unit charge

B: Energy       $2$: Rate of work-done

C: Voltage      $3$: Capacity to do work

a) A-$3$   B-$1$   C-$2$

b) A-$2$   B-$3$   C-$1$

c) A-$1$   B-$2$   C-$3$

d) A-$3$   B-$2$   C-$1$

Correct Answer: Option B

Solution:

How fast a work can be done is power.

How much work can be done is Energy.

How much work is done when moving an unit charge against an electric field is voltage.


Hence


A: Power     $2$: Rate of work-done 

B: Energy    $3$: Capacity to do work 

C: Voltage   $1$: Work-done in moving an unit charge

Circuit Theory

Match the following

A: Power        $1$: Work-done in moving an unit charge

B: Energy       $2$: Rate of work-done

C: Voltage      $3$: Capacity to do work

a) A-$3$   B-$1$   C-$2$

b) A-$2$   B-$3$   C-$1$

c) A-$1$   B-$2$   C-$3$

d) A-$3$   B-$2$   C-$1$

Correct Answer: Option B

Solution:

How fast a work can be done is power.
How much work can be done is Energy.
How much work is done when moving an unit charge against an electric field is voltage.

Hence
A: Power     $2$: Rate of work-done 

B: Energy    $3$: Capacity to do work 

C: Voltage   $1$: Work-done in moving an unit charge

Circuit Theory

Unit of energy is

a) Joules

b) watt-seconds

c) watt-hours

d) All the above

Correct Answer: Option D

Solution:

Energy is nothing but capability to do work. 

Since Energy = Work-done, its unit is also joules. 

Since Power = Rate of work done and $$P=W/t$$ $$P=E/t$$ $$E=Pt$$

Hence unit of energy is also watt-seconds or watt-hours

Circuit Theory

Unit of energy is

a) Joules

b) watt-seconds

c) watt-hours

d) All the above

Correct Answer: Option D

Solution:

Energy is nothing but capability to do work. 

Since Energy = Work-done, its unit is also joules. 

Since Power = Rate of work done and $$P=W/t$$ $$P=E/t$$ $$E=Pt$$

Hence unit of energy is also watt-seconds or watt-hours

Circuit Theory

One kWh is equal to

a) $860\;kcal$

b) $800\;kcal$

c) $746\;kcal$

d) $736\;kcal$

Correct Answer: Option A

Solution:

One kWh = $860.4\;kcal$ 

One calorie or kcal = $4184\;J$ or $4.18\;kJ$

Circuit Theory

One kWh is equal to

a) $860\;kcal$

b) $800\;kcal$

c) $746\;kcal$

d) $736\;kcal$

Correct Answer: Option A

Solution:

One kWh = $860.4\;kcal$ 

One calorie or kcal = $4184\;J$ or $4.18\;kJ$

Circuit Theory

A $100\;W$ bulb, a $40\;W$ fluorescent lamp and a $60\;W$ fan in a room operates for an average duration of $5$ hours in a day. What will be the total energy consumed during the month of April $2016$?

a) $3\;kWh$

b) $30\;kWh$

c) $300\;kWh$

d) $3000 \;kWh$

Correct Answer: Option B

Solution:

Total power = $100+40+60\;W$ $=200\;W$

Since Energy = Power $×$ time period

Power = $200\;W$

Time period = $5$ hours in $30$ days (April $2016$)

$$E=200×5×30\;Wh$$ $$E=30,000\;Wh$$ $$E=30\;kWh$$

Circuit Theory

A $100\;W$ bulb, a $40\;W$ fluorescent lamp and a $60\;W$ fan in a room operates for an average duration of $5$ hours in a day. What will be the total energy consumed during the month of April $2016$?

a) $3\;kWh$

b) $30\;kWh$

c) $300\;kWh$

d) $3000 \;kWh$

Correct Answer: Option B

Solution:

Total power = $100+40+60\;W$ $=200\;W$

Since Energy = Power $×$ time period

Power = $200\;W$

Time period = $5$ hours in $30$ days (April $2016$)

$$E=200×5×30\;Wh$$ $$E=30,000\;Wh$$ $$E=30\;kWh$$

Circuit Theory

Which of the following is NOT true regarding passive electronic components?

a) Passive components cannot control the flow of electric current through a circuit but can limit the flow of electric current.

b)  Passive components do not depend upon the external source of energy or voltage to perform a specific operation.

c) Passive components temporarily store the electrical energy in the form of static electric field or magnetic field.

d) Passive components can amplify or increase the power of electrical signal.

Correct Answer: Option D

Solution:

Passive components
1) cannot provide power gain or amplification $$P_{out} \leq P_{in}$$
2) it disspates power $$\int_{-inf}^t v(t)\; i(t)\; dt\;=\;+ve$$
3) It can store energy in the form of electric field or magnetic field for finite period of time only. (i.e) it cannot supply power for infinite period of time.
4) They do not need any external power source to perform its work.
5) It cannot control the current flow by means of another electrical signal.

Circuit Theory

Which of the following is NOT true regarding passive electronic components?

a) Passive components cannot control the flow of electric current through a circuit but can limit the flow of electric current.

b) 
Passive components do not depend upon the external source of energy or voltage to perform a specific operation.

c) Passive components temporarily store the electrical energy in the form of static electric field or magnetic field.

d) Passive components can amplify or increase the power of electrical signal.

Correct Answer: Option D

Solution:

Passive components
1) cannot provide power gain or amplification $$P_{out} \leq P_{in}$$
2) it disspates power $$\int_{-inf}^t v(t)\; i(t)\; dt\;=\;+ve$$
3)It can store energy in the form of electric field or magnetic field for finite period of time only. (i.e) it cannot supply power for infinite period of time.
4) They do not need any external power source to perform its work.
5) It cannot control the current flow by means of another electrical signal.