Electricity Test

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Electricity Test - 53

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Question 1
1 / -0

What is the highest equivalent resistance that can be secured by combinations of four coils of resistance $$4\ \Omega$$, $$8\ \Omega$$, $$12\ \Omega$$, $$24\ \Omega$$?

A
$$24\ \Omega$$

B
$$48\ \Omega$$

C
$$12\ \Omega$$

D
$$4\ \Omega$$

Solution

Connecting resistances in series with each other increases the equivalent resistance of the combination.
Let $$4\ \Omega$$, $$8\ \Omega$$, $$12\ \Omega$$, and $$24\ \Omega$$ resistors are connected in series.
$$\therefore$$ Equivalent resistance of the circuit
$$R_{eq} = 4+8+12+24$$
$$\Rightarrow R_{eq} = 48$$ ohm
Question 2
1 / -0

Find energy in kWh consumed in 10 hours by four devices of power $$500\ W$$ each?

A
$$10\ kWh$$

B
$$20\ kWh$$

C
$$30\ kWh$$

D
$$40\ kWh$$

Solution

Hint:
Power is the ratio of the work and the time. Energy consumed in four devices is four times the energy consumed in one device.

Step 1: Given data
Power consumed by each device, $$P=500\ W=0.50\ kW$$
Time, $$t=10\ h$$
Expression for the energy consumed is written as,
$$W=P\times t$$

Step 2: Calculations,
Energy consumed by each device, $$W=0.50 \times 10 = 5kWh$$
Energy consumed by four devices, $$ = 4 \times 5 = 20\ kWh$$
Question 3
1 / -0

A piece of wire of resistance $$R$$ is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is $$R'$$, then the ratio $$\dfrac{R}{R'}$$ is -

A
$$\dfrac{1}{25}$$

B
$$\dfrac{1}{5}$$

C
$$5$$

D
$$25$$

Solution

$$\textbf{Step 1: Resistance of each part after cutting} $$
We know that, Resistance $$R = \rho \dfrac{\ell}{A} $$
Since, $$R$$ is directly proportional to length of the wire.
Therefore the resistance of every part after cutting the wire in $$5$$ parts is
$$R_{1} = R_{2} = R_{3} = R_{4} = R_{5} = \dfrac{R}{5} $$

$$\textbf{Step 2: Equivalent Resistance}$$
When all the resistances are connected in parallel, let R' be the equivalent resistance
So, $$\dfrac{1}{R'} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} + \dfrac{1}{R_4} + \dfrac{1}{R_5}$$

$$\Rightarrow \dfrac{1}{R'} = \dfrac{5}{R} + \dfrac{5}{R} + \dfrac{5}{R} + \dfrac{5}{R} + \dfrac{5}{R} $$

$$\Rightarrow R' = \dfrac{R}{25} $$

$$\Rightarrow$$ $$ \dfrac{R}{R'} = 25$$

Hence, Option (D) is correct.
Question 4
1 / -0

An electric bulb is rated $$220\ V$$ and $$100\ W%$$. When it is operated on $$110\ V$$, the power consumed will be:

A
$$100\ W$$

B
$$75\ W$$

C
$$50\ W$$

D
$$25\ W$$

Solution

$$P=V^2/R$$

At $$220\ V,$$
$$100=220^2/R \quad ..........(i)$$

At $$110\ V,$$
$$P=110^2/R\quad ...........(ii) $$

Dividing (i) from (ii),
$$P/100=110^2/220^2$$
$$P=25\ W$$
Question 5
1 / -0

A 5 amp fuse wire can withstand a maximum power of 1 W in circuit. The resistance of the fuse wire is

A
$$0.2\ \Omega$$

B
$$5\ \Omega$$

C
$$0.4\ \Omega$$

D
$$0.04\ \Omega$$

Solution

Given: Power $$(P) = 1 \space W$$
Current $$(I) = 5 \space A$$
Resistance $$(R) = ?$$
The power of a resistor is the rate of flow of energy through it. Maximum power is the limiting energy flow after which the resistor (or wire here) would fuse.
We know that,
$$P=I^2R$$
$$\Rightarrow R=\dfrac{P}{I^2}=\dfrac{1W}{25A^2}=0.04\Omega$$
Question 6
1 / -0

If a third identical resistor is added in parallel in a circuit of two identical parallel resistors. Calculate the ratio of the new equivalent resistance to the old?

A
$$\dfrac{4}{9}$$

B
$$\dfrac{2}{3}$$

C
$$1$$

D
$$\dfrac{3}{2}$$

E
$$\dfrac{9}{4}$$

Solution

Let the resistance of each resistor be $$R$$.
Equivalent resistance of two resistors in parallel, $$\dfrac{1}{R_{eq}} = \dfrac{1}{R} + \dfrac{1}{R}$$
$$\implies R_{eq} = \dfrac{R}{2}$$

Equivalent resistance of three resstors in parallel, $$\dfrac{1}{R'_{eq}} = \dfrac{1}{R} + \dfrac{1}{R}+ \dfrac{1}{R}$$
$$\implies R'_{eq} = \dfrac{R}{3}$$

$$\therefore\ \dfrac{R'_{eq}}{R_{eq}} = \dfrac{R/3}{R/2} = \dfrac{2}{3}$$
Question 7
1 / -0

Three resistor of $$10 \Omega$$, $$20 \Omega$$, and $$30 \Omega$$ are connected in parallel in a circuit. Calculate the equivalent resistance.

A
$$6000 \ \Omega$$

B
$$0.017 \ \Omega$$

C
$$1 \ \Omega$$

D
$$60 \ \Omega$$

E
$$5.45 \ \Omega$$

Solution

Given : $$R_1 = 10\Omega$$ $$R_2= 20\Omega$$ $$R_3 = 30\Omega$$
Equivalent resistance of parallel combination $$\dfrac{1}{R_{eq}} = \dfrac{1}{R_1} +\dfrac{1}{R_2} + \dfrac{1}{R_3}$$

$$\therefore$$ $$\dfrac{1}{R_{eq}} = \dfrac{1}{10} +\dfrac{1}{20} + \dfrac{1}{30}$$ $$\implies R_{eq} = 5.45$$ $$\Omega$$
Question 8
1 / -0

In a DC circuit when a switch is closed to operate the circuit, the electrons that cause the current:

A
travel from the negative plate of the battery to the positive plate

B
travel from the positive plate of the battery to the negative plate

C
travel from one resistance to the next resistance

D
travel from the negative plate of the battery to the positive plate inside the battery

E
travel from one atom in the circuit to the next atom

Solution

In a DC circuit , when switch is closed , the free electrons in wire are drifted from negative plate of the battery to positive plate.
Question 9
1 / -0

A fuse wire with radius $$1\ mm$$ blows at $$1.5\ A$$. The radius of the fuse wire of the same material to blow at $$3\ A$$ will be:

A
$$\displaystyle { 3 }^{ { 1 }/{ 4 } }\ mm$$

B
$$\displaystyle { 4 }^{ { 1 }/{ 3 } }\ mm$$

C
$$\displaystyle { 3 }^{ { 1 }/{ 2 } }\ mm$$

D
$$\displaystyle { 2 }^{ { 1 }/{ 3 } }\ mm$$

Solution

For fusing current, its relation with radius is given as: $$I^2\propto r^3$$
$$\Rightarrow \left (\dfrac{I_2}{I_1}\right )^2=\left (\dfrac{r_2}{r_1}\right )^3$$

$$\Rightarrow r_2=r_1\left (\dfrac{I_2}{I_1}\right )^{2/3}$$

$$\Rightarrow r_2=1\left (\dfrac{3}{1.5}\right )^{2/3}$$

$$\Rightarrow r_2=\left (2\right )^{2/3}=4^{1/3}\ mm$$
Question 10
1 / -0

A current is flowing in a circuit consisting of a resistor and a battery.
What happens to the power dissipated in the resistor when the resistance is quadrupled and the voltage remains constant?

A
It is doubled

B
It is quadrupled

C
It is halved

D
It is quartered

E
It remains the same

Solution

Let the voltage of the battery be $$V$$ and initial resistance be $$R$$.
Thus, the initial power dissipated is $$P= \dfrac{V^2}{R}$$
Now the resistance is quadrupled keeping the voltage constant i.e $$R' = 4R$$
$$\therefore$$ New power dissipattion is $$P' = \dfrac{V^2}{R'} = \dfrac{V^2}{4R} = \dfrac{P}{4}$$
Hence the power dissipated is quartered.

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Electricity Test - 53

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Electricity Test - 53

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Question 1

$$24\ \Omega$$

$$48\ \Omega$$

$$12\ \Omega$$

$$4\ \Omega$$

Solution

Question 2

$$10\ kWh$$

$$20\ kWh$$

$$30\ kWh$$

$$40\ kWh$$

Solution

Question 3

$$\dfrac{1}{25}$$

$$\dfrac{1}{5}$$

$$5$$

$$25$$

Solution

Question 4

$$100\ W$$

$$75\ W$$

$$50\ W$$

$$25\ W$$

Solution

Question 5

$$0.2\ \Omega$$

$$5\ \Omega$$

$$0.4\ \Omega$$

$$0.04\ \Omega$$

Solution

Question 6

$$\dfrac{4}{9}$$

$$\dfrac{2}{3}$$

$$1$$

$$\dfrac{3}{2}$$

$$\dfrac{9}{4}$$

Solution

Question 7

$$6000 \ \Omega$$

$$0.017 \ \Omega$$

$$1 \ \Omega$$

$$60 \ \Omega$$

$$5.45 \ \Omega$$

Solution

Question 8

travel from the negative plate of the battery to the positive plate

travel from the positive plate of the battery to the negative plate

travel from one resistance to the next resistance

travel from the negative plate of the battery to the positive plate inside the battery

travel from one atom in the circuit to the next atom

Solution

Question 9

$$\displaystyle { 3 }^{ { 1 }/{ 4 } }\ mm$$

$$\displaystyle { 4 }^{ { 1 }/{ 3 } }\ mm$$

$$\displaystyle { 3 }^{ { 1 }/{ 2 } }\ mm$$

$$\displaystyle { 2 }^{ { 1 }/{ 3 } }\ mm$$

Solution

Question 10

It is doubled

It is quadrupled

It is halved

It is quartered

It remains the same

Solution

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