Electricity Test

Result

Electricity Test - 54

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

A wire of resistance R is elongated n-fold to make a new uniform wire. The resistance of new wire

A
$$nR$$

B
$$n^2R$$

C
$$2nR$$

D
$$2n^2R$$

Solution

We know that resistance $$R=\rho \dfrac{l}{A}$$, where $$\rho$$ is the resistivity of the material, $$l$$ is the length of the wire, and $$A$$ is the cross-sectional area of the wire.
The wire is elongated keeping the volume fixed.
$$\text{Old Volume} = \text{New volume}$$
$$\Rightarrow l\times A = l'\times A'$$
$$\Rightarrow l\times A=nl\times A'$$
$$\Rightarrow A'=\dfrac{A}{n}$$
Thus, the area is decreased by n times.
So, the altered resistance would be $$R'=\rho \dfrac{nl}{\dfrac{A}{n}}=n^2 \rho \dfrac{l}{A}=n^2R$$
Question 2
1 / -0

The resistance across $$A$$ and $$B$$ in the given figure will be:

A
$$3R$$

B
$$R$$

C
$$\cfrac{R}{3}$$

D
None of the above

Solution

Given: $$R_1 = R_2 = R_3 = R$$

The three resistances are connected in parallel between the points A and B.
Equivalent resistance for parallel combination:
$$\Rightarrow \dfrac{1}{R_{AB}} =\dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}$$

$$\Rightarrow \dfrac{1}{R_{AB}} =\dfrac{1}{R} + \dfrac{1}{R} + \dfrac{1}{R} = \dfrac{3}{R}$$

$$\Rightarrow $$ $$R_{AB} = \dfrac{R}{3}$$
Question 3
1 / -0

In shown diagram, if voltage of battery is doubled and resistance kept constant then what is the new Power dissipated at resistor(Consider $$P$$ was the initial power dissipation) ?

A
$$\dfrac{P}{4}$$

B
$$\dfrac{P}{2}$$

C
P

D
2P

E
4P

Solution

Power, $$P=\dfrac{V^2}{R}$$
As R is kept constant so $$P \propto V^2$$
$$P'$$is new power dissipation.
So, $$\dfrac{P'}{P}=\dfrac{(2V)^2}{V^2}=\dfrac{4V^2}{V^2}$$ or $$P'=4P$$
Thus, option E is correct.
Question 4
1 / -0

This question relates to the DC circuit shown below.
The values for current, voltage, and resistance in the circuit are graphed from point A through point G in the graphs directly below.
Which of the graphs shows the resistance from point A to point G?

A
A

B
B

C
C

D
D

E
E

Solution

Here the resistance from A to G through the circuit is increasing. Since the all resistances have same value so the resistance will be increased same amount at each point of the circuit from A to G. Here graph B shows the resistance increase by the same value from point to point. Thus, option B will be correct.
Question 5
1 / -0

All resistors in the circuit have the same resistance, $$R$$, and the battery is having a constant voltage of $$V$$.
Find out power dissipated by resistor $$f$$. Given that the power dissipated by resistor $$e$$ is $$P$$:

A
$$\dfrac{P}{6}$$

B
$$\dfrac{P}{3}$$

C
$$\dfrac{P}{2}$$

D
$$P$$

E
$$2P$$

Solution

The power dissipated by a resistor is $$i^2R$$
where $$i$$ is the current through the resistor of resistance $$R$$.
Since both $$i$$ and $$R$$ for the resistors $$e$$ and $$f$$ are equal (they are of equal resistance and connected in series), the power dissipated by both of them is the same and equal to $$P.$$
Question 6
1 / -0

Three resistors $$R_1, R_2$$, and $$R_3$$ are connected as shown below.
The resistance of $$R_3$$ is $$1 \Omega$$ and it is known that the value of $$R_3$$ is smaller than the values of $$R_1$$ or $$R_2$$. The resistances of $$R_1$$ and $$R_2$$ are unknown.
Select the best statement describing the equivalent resistance of this combination:

A
The equivalent resistance of this combination must be lower than $$1\Omega$$.

B
The equivalent resistance of this combination may be lower than $$1\Omega$$.

C
The equivalent resistance of this combination must be greater than $$1\Omega$$.

D
The equivalent resistance of this combination may be greater than $$1\Omega$$.

E
The equivalent resistance of this combination cannot be estimated without additional information.

Solution

The resistances in given diagram are in parallel combination , therefore equivalent resistance of the given combination ,
$$1/R=1/R_{1}+1/R_{2}+1/R_{3}$$ ,
it is clear from the equation that value of $$R$$ is less than $$R_{1}$$ or$$R_{2}$$ or $$R_{3} =1$$ , therefore equivalent resistance of this combination is less than 1 ohm .
Question 7
1 / -0

Metals are good conductor because

A
outer electrons are strongly bound to the atom.

B
outer electrons are loosely bound to the atom.

C
inner electrons are loosely bound to the atom.

D
protons can detach from the nucleus and conduct electricity.

Solution

We know that metals are good conductor because they have contain free electrons in the valence shell. Thus, the outer electrons are loosely bound to the atom.
Question 8
1 / -0

What is the power dissipated by each lamp of resistance $$60 \Omega$$?

A
1 W

B
2 W

C
6 W

D
12 W

Solution

Resistance are in series
So total resistance $$R_T=R+R=6+6=12$$
current, $$I=\dfrac{V}{R_T}=\dfrac{12}{12}=1A$$
Power dissipated across each reistance $$P=i^2R=1^2\times 6=6W$$
Question 9
1 / -0

The dry cell of E.M.F $$12\ V$$, is connected in series to a $$3 \Omega$$ and a $$6 \Omega$$ resistor. Calculate the total power supplied by the dry cell.

A
$$9\ W$$

B
$$72\ W$$

C
$$16\ W$$

D
$$2\ W$$

Solution

Equivalent resistance for resistance connected in series is given by

$$R_{eq} = R_1 +R_2$$

Since, in the given case $$3 \ \Omega$$ and $$6 \ \Omega$$ are connected in series.
Hence, equivalent resistance, $$R_{eq} =3+6=9 \ \Omega$$

Given, emf of cell $$V = 12 \ V$$

Hence, the power supplied by the cell is given by, $$P = \dfrac{V^2}{R}$$

$$P = \dfrac{12 \times 12}{9} = 16 \ W$$
Question 10
1 / -0

An air conditioner is rated 240 V, 1.5 kW. The air conditioner is switched on 8 hours each day. What is electrical energy consumed in 30 days?

A
2.88 kWh

B
360 kWh

C
120 kWh

D
240 kWh

Solution

Given,
Power of air conditioner $$P=1.5kW$$
Time to use per day $$t=8\ hr$$
Time to use in 30 days $$T=8\times 30\ hr$$

Energy consumed $$E=Power\times time =1.5\ kW\times 8\times 30\ hr=360kWh $$
Option B