Electricity Test

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Electricity Test - 55

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Weekly Quiz Competition

Question 1
1 / -0

Calculate the unknown resistance R of the circuit as shown in figure, all resistance are connected in the series. The current is flowing through circuit is 2A and battery is of 20 voltage.

A
$$1 \Omega$$

B
$$2 \Omega$$

C
$$4 \Omega$$

D
$$6 \Omega$$

E
$$12 \Omega$$

Solution

Given : $$V =20$$ volts $$I = 2$$ A
Total resistance of the circuit $$R_{eq} =R + 4+4 = R+8$$
Using $$V = IR_{eq}$$
$$\therefore$$ $$20 =2 (R+8)$$ $$\implies R = 2\Omega$$
Question 2
1 / -0

Three resistors ($$2, 5$$ and $$7$$ ohm) are wires as shown in the diagram. The equivalent resistance of this combination (in ohm) is:

A
$$\dfrac{59}{70}$$

B
$$\dfrac{70}{59}$$

C
$$\dfrac{1}{14}$$

D
$$14$$

E
Cannot be determined without additional information.

Solution

Given : $$R_1 = 2\Omega$$ $$R_2= 5\Omega$$ $$R_3 = 7\Omega$$
Equivalent resistance of parallel combination
$$\dfrac{1}{R_{eq}} =\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}$$
$$\therefore$$ $$\dfrac{1}{R_{eq}} =\dfrac{1}{2}+\dfrac{1}{5}+\dfrac{1}{7}$$
$$\Rightarrow \dfrac{1}{R_{eq}}=\dfrac{35+14+10}{70}$$
$$\Rightarrow R_{eq} =\dfrac{70}{59}$$ $$\Omega$$
Question 3
1 / -0

Two electrical bulbs have tungsten filament of same length. If one of them gives 60 watts and other 100 watts, then

A
100 watts bulb has thicker filament

B
60 watt bulb has thicker filament

C
Both filaments are of same thickness

D
It is impossible to get different wattage unless lengths are different

Solution

power $$P=\dfrac{V^2}{R}$$, where $$V$$ is the voltage and $$R$$ is the resistance.
So for higher power, the resistance should be less.
Therefore, the resistance of $$100\ W$$ bulb should be less.
We know that resistance $$R=\dfrac{\rho l}{A}$$, where $$\rho$$ is the resistivity, $$l$$ is the length, and $$A$$ is the area.
when the resistivity and the length are fixed, the area should be more for less resistance.
Thus, the $$100\ W$$ bulb should have a thicker filament.
Question 4
1 / -0

The resistors of resistances $$2\Omega, 4\Omega, 5\Omega$$ are connected in parallel. The total resistance of the combination will be:

A
$$\dfrac {29}{10}\Omega$$

B
$$\dfrac {19}{20}\Omega$$

C
$$\dfrac {10}{20}\Omega$$

D
$$\dfrac {20}{19}\Omega$$

Solution

Given: $$R_1=2\ \Omega$$, $$R_2=4\ \Omega$$, $$R_3=5\ \Omega$$

In parallel combination-
$$\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}$$

$$\Rightarrow\dfrac{1}{R}=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{5}$$

$$\Rightarrow R=\dfrac{20}{19}\Omega$$
Question 5
1 / -0

If four resistances, each of value $$1\ ohm$$, are connected in series, what will be the resultant resistance?

A
3 ohm

B
4 ohm

C
5 ohm

D
2 ohm

Solution

In series combination , equivalent resistance is given by-

$$R=R_1+R_2+R_3+R_4+R_5+.....$$

Here there are four resistors.

$$R=1+1+1+1$$

$$\implies R=4\Omega$$

Answer-(B)
Question 6
1 / -0

Calculate the equivalent resistance when two resistances of $$3\ \Omega$$ and $$6\ \Omega$$ are connected in parallel.

A
$$1 \ \Omega$$

B
$$3 \ \Omega$$

C
$$2 \ \Omega$$

D
$$6 \ \Omega$$

Solution

We know that in parallel combination, equivalent resistance is given by-

$$\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}$$

$$\implies \dfrac{1}{R}=\dfrac{1}{3}+\dfrac{1}{6}$$

$$\implies \dfrac{1}{R}=\dfrac{2+1}{6}=\dfrac{1}{2}$$

$$\implies R=2\ \Omega$$

Answer-(C)
Question 7
1 / -0

Two resistances are connected in parallel and a current is sent through the combination. The current divides itself:

A
In the inverse ratio of resistance

B
In the direct ratio of resistance

C
Equally in both the resistance

D
In none of the above manner

Solution

Current flows in inverse ratio of resistance
$$\dfrac{i_1}{i_2}=\dfrac{R_2}{R_1}$$
Question 8
1 / -0

Electrical power P is given by the expression P = $$\dfrac{Q \times V}{t}$$. time. What does $$Q$$ and $$V$$ stand for?

A
Charge, energy

B
Current, voltage

C
Current, power

D
Charge, voltage

Solution

$$Q$$ is the symbol for charge. $$V$$ is the symbol for voltage.
Question 9
1 / -0

What do you add to distilled water for making it to conduct electricity.

A
sulphuric acid

B
water

C
milk

D
base

Solution
Question 10
1 / -0

Two resistance are connected in series as shown in diagram. What is the current through the $$5\ ohm$$ resistance?

A
$$2 A$$

B
$$3 A$$

C
$$4 A$$

D
$$5 A$$

Solution

Voltage across $$5\Omega$$ resistor is $$V=10V$$

Current, $$I=\dfrac{V}{R}=\dfrac{10}{5}$$

$$\implies I=2A$$

Answer-(A)

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Electricity Test - 55

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Electricity Test - 55

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SHARING IS CARING

Question 1

$$1 \Omega$$

$$2 \Omega$$

$$4 \Omega$$

$$6 \Omega$$

$$12 \Omega$$

Solution

Question 2

$$\dfrac{59}{70}$$

$$\dfrac{70}{59}$$

$$\dfrac{1}{14}$$

$$14$$

Cannot be determined without additional information.

Solution

Question 3

100 watts bulb has thicker filament

60 watt bulb has thicker filament

Both filaments are of same thickness

It is impossible to get different wattage unless lengths are different

Solution

Question 4

$$\dfrac {29}{10}\Omega$$

$$\dfrac {19}{20}\Omega$$

$$\dfrac {10}{20}\Omega$$

$$\dfrac {20}{19}\Omega$$

Solution

Question 5

3 ohm

4 ohm

5 ohm

2 ohm

Solution

Question 6

$$1 \ \Omega$$

$$3 \ \Omega$$

$$2 \ \Omega$$

$$6 \ \Omega$$

Solution

Question 7

In the inverse ratio of resistance

In the direct ratio of resistance

Equally in both the resistance

In none of the above manner

Solution

Question 8

Charge, energy

Current, voltage

Current, power

Charge, voltage

Solution

Question 9

sulphuric acid

water

milk

base

Solution

Question 10

$$2 A$$

$$3 A$$

$$4 A$$

$$5 A$$

Solution

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