Electricity Test

Result

Electricity Test - 51

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

Two metallic wires of $$\displaystyle 6\Omega $$ and $$\displaystyle 3\Omega $$ are in connection. What will be the mode of connection so that to get effective resistance of $$\displaystyle 2\Omega $$ ?

A
Parallel

B
Series

C
Both

D
None

Solution

$$R_1$$=$$6\Omega$$ , $$R_2$$=$$3\Omega$$
Case (1): In Series connection :
$$R_{eq}$$=$$R_1$$ + $$R_2$$
$$R_{eq}$$=$$6$$ + $$3$$
$$R_{eq}$$=$$9\Omega$$
So series connection not possible.

Case (2): In Parallel connection :
$$\frac{1}{R_{eq}}$$=$$\frac{1}{R_1}+\frac{1}{R_2}$$
$$\frac{1}{R_{eq}}$$=$$\frac{1}{6}+\frac{1}{3}$$
$$R_{eq}$$ = $$\frac{6\times3}{6+3}$$= $$\frac{18}{9}$$=$$2\Omega$$
Hence the two resistors are connected in parallel.
Question 2
1 / -0

Three resistors each having the same resistance are connected in parallel. Their equivalent resistance is $$\displaystyle 5\ \Omega $$. If they are connected in series, their equivalent resistance is.

A
$$\displaystyle 3\ \Omega $$

B
$$\displaystyle 45\ \Omega $$

C
$$\displaystyle 5\ \Omega $$

D
$$\displaystyle 8\ \Omega $$

Solution
Question 3
1 / -0

Number of KWh in 1 Joule.

A
$$\displaystyle 3.6\times { 10 }^{ 6 }\ KWh$$

B
$$\displaystyle 2.77\times { 10 }^{ -7 }\ KWh$$

C
$$\displaystyle 600\ KWh$$

D
$$\displaystyle 1.6\times { 10 }^{ -19 }\ KWh$$

Solution

We know that $$KWh$$ is kilo watt-hour-
In SI unit-
$$K = 10^3$$
$$hour = 3600\ s$$
$$1\ kWh = 3.6\times 10^6\ J$$
$$\therefore$$ $$1$$ $$J = \dfrac{1}{3.6\times 10^6} = 2.77\times 10^{-7}$$ $$kWh$$
Question 4
1 / -0

In the circuit, $$\displaystyle 5\Omega $$ resistor develops 45J/s due to current flowing through it. The power developed per second across $$\displaystyle 12\Omega $$ resistor is:

A
16W

B
192W

C
36W

D
64W

Solution

Voltage across $$5\Omega$$ $$V = \sqrt{PR} = \sqrt{45\times 5} = 15$$ volts
$$\therefore$$ Current $$I_1 = \dfrac{15}{5} = 3$$ A
Equivalent resistance of series combination of $$3\Omega$$ and $$6\Omega$$ $$R' = 9+6 = 15\Omega$$
$$\therefore$$ Current $$I_2 = \dfrac{15}{15} = 1$$ A
Thus total current flowing through $$12\Omega$$ $$I = I_1+I_2 = 3+1 = 4$$ A
$$\therefore$$ Power developed across $$12\Omega$$ $$P = I^2 (12) = 4^2\times 12 = 192$$ W
Question 5
1 / -0

An electricity bulb of 100 watt is connected to supply of electricity of 220V. Resistance of filament is

A
$$\displaystyle 484\Omega $$

B
$$\displaystyle 100\Omega $$

C
$$\displaystyle 22000\Omega $$

D
$$\displaystyle 242\Omega $$

Solution

Given : $$V =220$$ volts $$P =100$$ W
$$\therefore$$ Resistance of the filament $$R = \dfrac{V^2}{R}$$
$$\implies$$ $$R = \dfrac{220^2}{100} = 484\Omega$$
Question 6
1 / -0

When two resistances $$\displaystyle { R }_{ 1 }$$ and $$\displaystyle { R }_{ 2 }$$ are connected in series, they consume 12W power. When they are connected in parallel, they consume 50W power. What is the ratio of the powers of $$\displaystyle { R }_{ 1 }$$ and $$\displaystyle { R }_{ 2 }$$

A
$$1/4$$

B
$$4$$

C
$$3/2$$

D
$$3$$

Solution

Let the powers of the resistances be $$P_1$$ and $$P_2$$.
Parallel connection : $$P_{eq} = P_1+P_2$$
$$\therefore$$ $$50 = P_1+P_2$$ $$\implies P_2 = 50-P_1$$
Series connection : $$P_{eq} = \dfrac{P_1P_2}{P_1+P_2}$$
$$\therefore$$ $$12 = \dfrac{P_1(50 - P_1)}{50}$$
OR $$P_1^2 - 50P_1 + 600 = 0$$ $$\implies$$ $$P_1 = 30$$ W
$$\implies$$ $$P_2 = 50 - 30 = 20$$ W
Thus ratio of powers $$\dfrac{P_1}{P_2} = \dfrac{3}{2}$$
Question 7
1 / -0

How much current is drawn by the motor of 1 H.P. from 220 volt supply.

A
3.4A

B
2A

C
6A

D
22A

Solution

We know that $$1$$ H.P $$ = 746$$ watt
Current $$I = \dfrac{P}{V} = \dfrac{746}{220} = 3.4$$ A
Question 8
1 / -0

A bulb of $$22\ \Omega $$ is producing light when connected to a $$220\ V$$ supply. What is the electric power of the bulb?

A
$$2200\ W$$

B
$$1000\ W$$

C
$$22\ W$$

D
$$220\ W$$

Solution

Given,
Resistance of the bulb, $$R = 22\Omega$$
Voltage, $$V = 220$$ volts
We know,
Power of the bulb, $$P = \dfrac{V^2}{R}$$

$$P = \dfrac{220^2}{22}$$

$$P =2200\ W$$
Question 9
1 / -0

An electric bulb of resistance of $$1000 \Omega$$ draws a current of $$0.05 A$$. Calculate the power of the bulb :

A
$$500 \ W$$

B
$$25 \ kW$$

C
$$250 \ W$$

D
$$650 \ W$$

Solution

Given: $$Resistance(R) = 1000 \Omega, Current(I) = 0.5 A$$
We know that, Power $$P = {I}^{2} \cdot R = {\left(0.5 A\right)}^{2} \times \left(1000\right) \Omega = 250 W$$.
Question 10
1 / -0

An electric bulb of resistance $$20 \Omega$$ draws a current of $$0.04 A$$. Calculate the potential difference at the ends.

A
$$0.8 V$$

B
$$2 V$$

C
$$0.032 V$$

D
$$5 V$$

Solution

Given,
Resistance of resistor $$R = 20 \Omega$$
Current draws $$ I = 0.04 A$$
According to ohm's law
Potential difference, $$V = IR = 0.04 \times 20 = 0.8 V$$.
Hence Option A

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Electricity Test - 51

Result

Electricity Test - 51

Score

-

Rank

-

SHARING IS CARING

Question 1

Parallel

Series

Both

None

Solution

Question 2

$$\displaystyle 3\ \Omega $$

$$\displaystyle 45\ \Omega $$

$$\displaystyle 5\ \Omega $$

$$\displaystyle 8\ \Omega $$

Solution

Question 3

$$\displaystyle 3.6\times { 10 }^{ 6 }\ KWh$$

$$\displaystyle 2.77\times { 10 }^{ -7 }\ KWh$$

$$\displaystyle 600\ KWh$$

$$\displaystyle 1.6\times { 10 }^{ -19 }\ KWh$$

Solution

Question 4

16W

192W

36W

64W

Solution

Question 5

$$\displaystyle 484\Omega $$

$$\displaystyle 100\Omega $$

$$\displaystyle 22000\Omega $$

$$\displaystyle 242\Omega $$

Solution

Question 6

$$1/4$$

$$4$$

$$3/2$$

$$3$$

Solution

Question 7

3.4A

2A

6A

22A

Solution

Question 8

$$2200\ W$$

$$1000\ W$$

$$22\ W$$

$$220\ W$$

Solution

Question 9

$$500 \ W$$

$$25 \ kW$$

$$250 \ W$$

$$650 \ W$$

Solution

Question 10

$$0.8 V$$

$$2 V$$

$$0.032 V$$

$$5 V$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.