Electricity Test

Result

Electricity Test - 41

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Find out the total power supplied by the 60-V source in the circuit shown in figure.

A
360 W

B
200 W

C
400 W

D
180 W

Solution

First we will find the equivalent resistance of the circuit.
The $$7\Omega$$ and $$5\Omega$$ are in series so equivalent is $$7+5=12\Omega$$.
$$12\Omega$$ and $$6\Omega$$ are in parallel, so equivalent is $$12\times6/(12+6)=4\Omega$$.
These $$4\Omega$$ and $$12\Omega$$ are in parallel, so their equivalent is $$12\times4/(12+4)=3\Omega$$.
This $$3\Omega$$ and $$7\Omega$$ are in series so equivalent is $$3+7=10\Omega$$.
So the equivalent resistance of the circuit is $$R_{eq}=10\Omega$$.

Thus the power supplied is $$\dfrac{V^2}{R_{eq}}=60^2/10=360\ W$$
Question 2
1 / -0

Two resistors of resistance $$4\, \Omega$$ and $$6\, \Omega$$ are connected in parallel to a cell to draw $$0.5 A$$ current from the cell. What is the current through resistor $$6\, \Omega$$?

A
$$0.2 \ A$$

B
$$0.5 \ A$$

C
$$2.4 \ A$$

D
$$1.2 \ A$$

Solution

Two resistors of resistance $$4 \ Ω$$ and $$6 \ Ω$$ are connected in parallel.

Hence, equivalent resistance in parallel is given by,
$$\dfrac{1}{R_{eq}} = \dfrac{1}{R_1} + \dfrac{1}{R_2}$$
$$\Rightarrow \dfrac{1}{R_{eq}} = \dfrac{1}{4} + \dfrac{1}{6} = \dfrac{10}{24}$$
$$\Rightarrow R_{eq} = \dfrac{24}{10}= 2.4 \ \Omega$$

Using Ohm's law, $$V=iR$$
$$V=0.5\times2.4=1.2 \ V$$

Since, value of voltage is same for objects connected in parallel, Hence, voltage drop across $$6 \ \Omega$$ resistor is also $$1.2\ V$$

Hence, using Ohm's law , Current through resistor $$6 \ Ω$$:
$$1.2=i\times6$$
$$i=0.2 \ A$$
Question 3
1 / -0

The figure shows two bulbs with a resistance of $$10\ \Omega$$ with switches and a fuse connected to mains through a socket. How are the two bulbs joined and what is net resistance?

A
In parallel, $$20\ \Omega$$

B
in series, $$20\ \Omega$$

C
In parallel, $$5\ \Omega$$

D
In series , $$5\ \Omega$$

Solution

The two bulbs are connected in parallel.
Given,
Resistance of each bulb, $$R_1=R_2=R=10\ \Omega$$
For two resistors connected in parallel
Equivalent resistance, $$R_{eq}$$
$$\dfrac{1}{R_{eq}}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}$$

$$\dfrac{1}{R_{eq}}=\dfrac{1}{10}+\dfrac{1}{10}$$

$$\dfrac{1}{R_{eq}}=\dfrac{2}{10}=\dfrac 15$$

$$R_{eq}=5\ \Omega$$
Question 4
1 / -0

A wire of uniform thickness with a resistance of $$27\, \Omega$$ is cut into three equal pieces and they are joined in parallel. The equivalent resistance of the parallel combination will be :

A
$$3\, \Omega$$

B
$$9\, \Omega$$

C
$$27\, \Omega$$

D
$$49\, \Omega$$

Solution

In this case, a wire of uniform thickness with a resistance of 27 is cut into three equal pieces and they are joined in parallel. As we know that the resistance is directly proportional to the length of the wire, $$R\propto L$$, so when the wire is cut into 3, the resistance of the cut wire is reduced to one third of the total resistance. That is, the resistance of each cut wire becomes 27/3= 9 ohms.
Now, when these 3 wires are connected in a parallel connection, the total resistance is given as $$\frac { 1 }{ 9 } +\frac { 1 }{ 9 } +\frac { 1 }{ 9 } =\frac { 1 }{ 0.333 } =\quad 3\quad ohms$$.
Hence, the total resistance is now 3 ohms only.
Question 5
1 / -0

What is the current through an electrical appliance of rating 5 kW, 200 V and can you use a fuse which is rated 8 A ?

A
25 A, yes

B
2.5 A, yes

C
25 A, no

D
2.5 A, no

Solution

Fuses are safety devices that are to be built into our electrical system. If there were no fuses and we operated too many appliances on a single circuit, the cable carrying the power for that circuit would get extremely hot, short circuit, and possibly start a fire. To prevent electrical overloads, fuses are designed to trip or blow, stopping the flow of current to the overloaded cable.
The fuse must always be connected to the mains and it must be of correct value. For example, a 15-ampere fuse should trip when the current through it exceeds 15 amperes. A 20-ampere fuse should blow when the current through it exceeds 20 amps.
In this case, let us consider a device of power 5 kW, that is, 5000 W and 200 V.

The current through the appliance is given as $$I=P/V.$$

That is, $$I=\dfrac { P }{ V } =\dfrac { 5000 }{ 200 } =25A$$.

The fuse of 8 A cannot be used because the current through the appliance is
greater the fuse rating. When the 25 A current flows through the fuse it will
blow off and it will not be able to fulfill it purpose.

Hence, a fuse is rated 8 A cannot be used with this appliance.
Question 6
1 / -0

A particular resistance wire has a resistance of 3.0 ohm per metre. The total resistance of three lengths of this wire each 1.5 m long, joined in parallel will be :

A
$$2.5\, \Omega$$

B
$$3\, \Omega$$

C
$$4.5\, \Omega$$

D
$$1.5\, \Omega$$

Solution

Components connected in parallel are connected so the same voltage is applied to each component. In a parallel circuit, the voltage across each of the components is the same, and the total current is the sum of the currents through each component. In parallel circuits, each light has its own circuit, so all but one light could be burned out, and the last one will still function. A circuit composed solely of components connected in parallel is known as a parallel circuit.

To find the total resistance of all components, the reciprocals of the resistances of each component is added and the reciprocal of the sum is taken.

Total resistance will always be less than the value of the smallest resistance That is, $$\dfrac { 1 }{ { R }_{ total } } =\dfrac { 1 }{ { R }_{ 1 } } +\dfrac { 1 }{ { R }_{ 2 } } +\dfrac { 1 }{ { R }_{ 3 } } $$

In this case, the resistance of one meter wire is 3 ohms. So the resistance of 1.5 meters of the same wire is given as $$\dfrac { 3\times 1.5 }{ 1 } =4.5\quad ohms$$.

When three such wires of 1.5 meters length and resistance of 4.5 ohms are joined in parallel the resistance is given as $$\dfrac { 1 }{ { R }_{ total } } =\dfrac { 1 }{ 4.5 } +\dfrac { 1 }{ 4.5 } +\dfrac { 1 }{ 4.5 } \quad =\dfrac { 1 }{ 0.666 } =1.5\quad ohms$$.

Hence, the total resistance is given as 1.5 ohms.
Question 7
1 / -0

What we used to protect the electric circuits from over loading and short circuit ?

A
capacitance

B
inductance

C
fuse wire

D
filament wire.

Solution

A fuse is the device to protect the electric circuits from over loading and short circuit.
Fuses are safety devices that are to be built into our electrical system. If there were no fuses and we operated too many appliances on a single circuit, the cable carrying the power for that circuit would get extremely hot, short circuit, and possibly start a fire. To prevent electrical overloads, fuses are designed to trip or blow, stopping the flow of current to the overloaded cable.
Fuse is a piece of wire of a material with a very low melting point. When a high current flows through the circuit due to overloading or short circuit, the wires gets heated and melts. As a result, the circuit is broken and current stops flowing.
Hence, the fuse element is made of zinc, copper, silver, aluminum, or alloys to provide stable and predictable characteristics.
The fuse must always be connected to the mains and it must be of correct value. For example, a 15-ampere fuse should trip when the current through it exceeds 15 amperes. A 20-ampere fuse should blow when the current through it exceeds 20 amps. Hence correct option is C.
Question 8
1 / -0

Three resistors of $$6.0 \Omega, \, 2.0 \Omega$$ and $$4.0 \Omega$$ are joined to an ammeter A and a cell of e.m.f. 6.0 V as shown in figure. The effective resistance of the circuit is :

A
$$6.0\, \Omega$$

B
$$2.0\, \Omega$$

C
$$3.0\, \Omega$$

D
$$12.0\, \Omega$$

Solution

Components connected in series are connected along a single path, so the same current flows through all of the components. The current through each of the components is the same, and the voltage across the circuit is the sum of the voltages across each component. In a series circuit, every device must function for the circuit to be complete. One bulb burning out in a series circuit breaks the circuit. A circuit composed solely of components connected in series is known as a series circuit.The total resistance of resistors in series is equal to the sum of their individual resistances. That is, $${ R }_{ total }={ R }_{ 1 }+{ R }_{ 2 }+{ R }_{ 3 }$$.
Components connected in parallel are connected so the same voltage is applied to each component. In a parallel circuit, the voltage across each of the components is the same, and the total current is the sum of the currents through each component. In parallel circuits, each light has its own circuit, so all but one light could be burned out, and the last one will still function. A circuit composed solely of components connected in parallel is known as a parallel circuit.To find the total resistance of all components, the reciprocals of the resistances of each component is added and the reciprocal of the sum is taken. Total resistance will always be less than the value of the smallest resistance That is, $$\dfrac { 1 }{ { R }_{ total } } =\dfrac { 1 }{ { R }_{ 1 } } +\dfrac { 1 }{ { R }_{ 2 } } +\dfrac { 1 }{ { R }_{ 3 } } \quad that\quad is,\quad R={ { R }_{ total } }^{ -1 }$$.
In this case, the resistance $$R_2$$ and $$R_3$$ are connected in series and the combined resistance is connected in parallel with the resistance $$R_1.$$
Therefore, the combined resistance of $$R_2$$ and $$R_3$$ is given as $$4+2 = 6\Omega $$. The total resistance in the circuit is given as
$$\dfrac{ 1 }{ 6 } +\dfrac{ 1 }{ 6 } =\dfrac{ 1 }{ 0.33 } \ =\ 3\ \Omega$$.

Hence, the effective resistance of the circuit is 3 Ohms.
Question 9
1 / -0

The rating of a fuse connected in the general household lighting circuit is:

A
15 A

B
5 A

C
10 A

D
zero

Solution

General household wire has a current supply of 5 ampere. Hence fuse also have that. hence The rating of a fuse connected in the general household lighting circuit is $$5A$$
Question 10
1 / -0

Two resistors of resistance $$4\, \Omega$$ and $$6\, \Omega$$ are connected in parallel to a cell to draw 0.5 A current from the cell. What is current through resistor $$4\, \Omega$$ ?

A
0.3 A

B
0.5 A

C
0.2 A

D
1.2 A

Solution

In a parallel resistance system current in 1st branch =
$${ I }_{ 1 }=\left( \frac { { R }_{ 2 } }{ { R }_{ 1 }+{ R }_{ 2 } } \right) \times I\\$$
$${ I }_{ 1 }=\frac { 6 }{ 4+6 } \times 0.5\\$$
$${ I }_{ 1 }=\frac { 3 }{ 10 } =0.3A$$