Current Electricity Test

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Current Electricity Test - 30

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Weekly Quiz Competition

Question 1
1 / -0

For a heater rated at $$4 \ kW$$; $$220 \ V$$ . Calculate the resistance of the heater element.

A
$$20 \Omega$$

B
$$21.6\Omega$$

C
$$200\Omega$$

D
$$12.2\Omega$$

Solution

Given: $$V =220$$ volts $$P = 4000$$ W
Electric power is given by
$$P=\dfrac{V^2}{R}$$ where $$R$$ is resistance and $$V$$ is voltage.

$$\therefore$$ Resistance of the heater $$R = \dfrac{V^2}{P} =\dfrac{220^2}{4000} = 12.1\Omega$$
Question 2
1 / -0

A $$10\ k\Omega$$ resistor can be obtained by using

A
$$3\ k\Omega,5\ k\Omega,2\ k\Omega$$ resistors in series

B
$$3\ k\Omega,5\ k\Omega,2\ k\Omega$$ Resistors in parallel

C
$$3\ k\Omega,5\ k\Omega$$ in series and combination in parallel to $$2\ k\Omega$$ resistor in series

D
$$5\ k\Omega,5\ k\Omega,2\ k\Omega,$$ Resistors in series

Solution

When three resistors are connected in series, the equivalent resistance is $$R_{eq}=R_1+R_2+R_3$$.
In the question, $$R_1=3\ k\Omega$$, $$R_2=5\ k\Omega$$, $$R_3=2\ k\Omega$$.
$$\therefore R_{eq}=3+5+2 = 10\ k\Omega$$
Question 3
1 / -0

For a metallic wire, the ratio $$\displaystyle \frac{V}{i}$$ (where, V = applied potential difference and i = current flowing)

A
is independent of temperature

B
increases as the temperature rises

C
decreases as the temperature rises

D
increases or decreases as temperature rises depending upon the metal.

Solution

Metallic wires contain free electrons which help in conduction of electricity. As temperature increases, atoms start vibrating more vigourously, thereby increasing the number of collisions. These collisions hinder the movement of free electrons. Therefore, the resistance to movement of electrons is higher at higher temperatures. So, the ratio increases as temperature increases.
Question 4
1 / -0

An experiment to verify ohm's law, a conductor of resistance R is taken. During the experiment, temperature of the conductor increases with the flow of current. Resistance of the conductor will

A
Remains same

B
Decrease

C
Increase

D
First increases then decreases

Solution

Resistance of conductor $$R_T = R_o(1+\alpha \Delta T)$$ where $$\alpha>0$$
Thus resistance of the conductor increases with increase in temperature.
Question 5
1 / -0

A bulb uses 5000J of energy in 10s, calculate its power.

A
50 $$J/s$$

B
500 $$Watts$$

C
5 $$Watts$$

D
600 $$J/s$$

Solution

Given : $$E = 5000$$ J $$t = 10$$ s
Power of the bulb $$Power=\dfrac{Energy \quad spent}{Time}$$ $$P = \dfrac{E}{t} = \dfrac{5000}{10} = 500$$ $$Watts$$
Question 6
1 / -0

When does two elements are said to be in series?

A
When same current physically flows through both elements

B
When different current physically flows through both elements

C
When no current physically flows through both elements

D
None of these

Solution

When two resistors are connected end to end are said to be connected in series. In a series connection, the same current physically flows through both elements.
Question 7
1 / -0

What is the equivalent resistance of three $$1000 \ \Omega$$ resistors in series?

A
$$1000 \ \Omega$$

B
$$3000 \ \Omega$$

C
$$2000 \ \Omega$$

D
$$1100 \ \Omega$$

Solution

Given : $$3$$ resistors of resistance $$R = 1000 \ \Omega$$

Equivalent resistance in series combinationis given by,
$$R_{eq} = R_1+R_2+R_3$$

$$R_{eq} = R+R+R = 3R = 3000 \ \Omega$$
Question 8
1 / -0

Electromotive Force (EMF) may be defined as:

A
work done per coulomb on the charge.

B
drift velocity of electrons.

C
flow of electron.

D
the number of coulombs of charge per second.

Solution

Electromotive Force (EMF) may be defined as work done per coulomb on the charge.
Question 9
1 / -0

What is the total resistance when four resistors of $$20, 40, 60$$, and $$80$$ respectively are connected in parallel (in ohm)?

A
$$4.4$$

B
$$9.6$$

C
$$14.8$$

D
$$20$$

E
$$25.2$$

Solution

Given : $$R_1 =20\Omega$$ $$R_2 =40\Omega$$ $$R_3 =60\Omega$$ $$R_4 =80\Omega$$
Equivalent resistance for parallel combination $$\dfrac{1}{R_{eq}} = \dfrac{1}{R_1} +\dfrac{1}{R_2}+\dfrac{1}{R_3}+\dfrac{1}{R_4}$$
$$\therefore$$ $$\dfrac{1}{R_{eq}} = \dfrac{1}{20} +\dfrac{1}{40}+\dfrac{1}{60}+\dfrac{1}{80}$$ $$\implies R_{eq} =9.6\Omega$$
Question 10
1 / -0

Identify the changes in a circuit on adding a light bulb in parallel to the actual resistance of the circuit. It will:

A
Decrease the total resistance

B
Increase the total resistance

C
Make the voltage lost in each light bulb different

D
Make the current through each light bulb the same

E
Not change the total current through the circuit

Solution

For the parallel combination of two resistances

$$\dfrac{1}{R_{eq}} =\dfrac{1}{R_1} +\dfrac{1}{R_2}$$

$$\Rightarrow R_{eq}< min {R_1, R_2}$$
Light bulb has its own resistance and hence the total resistance of the circuit decreases when it is connected in parallel to the actual resistance of the circuit.

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Current Electricity Test - 30

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Current Electricity Test - 30

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SHARING IS CARING

Question 1

$$20 \Omega$$

$$21.6\Omega$$

$$200\Omega$$

$$12.2\Omega$$

Solution

Question 2

$$3\ k\Omega,5\ k\Omega,2\ k\Omega$$ resistors in series

$$3\ k\Omega,5\ k\Omega,2\ k\Omega$$ Resistors in parallel

$$3\ k\Omega,5\ k\Omega$$ in series and combination in parallel to $$2\ k\Omega$$ resistor in series

$$5\ k\Omega,5\ k\Omega,2\ k\Omega,$$ Resistors in series

Solution

Question 3

is independent of temperature

increases as the temperature rises

decreases as the temperature rises

increases or decreases as temperature rises depending upon the metal.

Solution

Question 4

Remains same

Decrease

Increase

First increases then decreases

Solution

Question 5

50 $$J/s$$

500 $$Watts$$

5 $$Watts$$

600 $$J/s$$

Solution

Question 6

When same current physically flows through both elements

When different current physically flows through both elements

When no current physically flows through both elements

None of these

Solution

Question 7

$$1000 \ \Omega$$

$$3000 \ \Omega$$

$$2000 \ \Omega$$

$$1100 \ \Omega$$

Solution

Question 8

work done per coulomb on the charge.

drift velocity of electrons.

flow of electron.

the number of coulombs of charge per second.

Solution

Question 9

$$4.4$$

$$9.6$$

$$14.8$$

$$20$$

$$25.2$$

Solution

Question 10

Decrease the total resistance

Increase the total resistance

Make the voltage lost in each light bulb different

Make the current through each light bulb the same

Not change the total current through the circuit

Solution

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