Electricity Test

Result

Electricity Test - 18

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

In the following figure, the equivalent resistance between the points $$A$$ and $$B$$ in ohm will be :

A
$$1$$

B
$$2$$

C
$$3$$

D
$$9$$

Solution

Resistance of each resistor $$R=3\Omega$$
According to above fig. Resistors are in parallel combination.
$$\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}$$

$${ R }_{ eq. }=\dfrac { R }{ 3 } =\dfrac { 3 }{ 3 } =1\Omega$$
Question 2
1 / -0

On a bulb is written 220 Volt and 60 watt. Find out the resistance of the bulb and the value of the current flowing through it.

A
806.66 ohm 0.27 ampere

B
500 ohm 12 ampere

C
200 ohm 14 ampere

D
100 ohm 11 ampere

Solution

$$P=\dfrac{V^2}{R}=VI$$

Hence, $$R=\dfrac{V^2}{P}=\dfrac{220\times 220}{60}$$

$$\implies R=806.66\Omega$$

$$I=\dfrac{P}{V}=\dfrac{60}{220}$$

$$\implies I=0.27A$$

Answer-(A)
Question 3
1 / -0

In the circuit shown in figure, the current flowing through $$ 5 \, \Omega $$ resistance is

A
$$0.5 \, A$$

B
$$0.6 \, A$$

C
$$0.9 \, A$$

D
$$1.5 \, A$$

Solution

$$\textbf{Step 1: Apply Kirchhoff's current law [Refer Fig.]} $$
Apply Kirchhoff's current law at junction A,
$$2.1 A = I_{1} + I_{2} $$ $$....(1)$$

$$\textbf{Step 2: Apply Kirchhoff's Voltage law } $$
Apply KVL in the loop (in clockwise direction)
$$8I_{1} + 2I_{1} - 5I_2-20I_{2} = 0 $$
$$\Rightarrow 10I_{1} = 25 I_{2} $$
$$\Rightarrow I_{1} = 2.5 I_{2} $$ $$....(2)$$

$$\textbf{Step 3: Solving equations} $$
Equation $$(1)$$ and $$(2)\Rightarrow $$$$2.1 = 2.5 I_{2} +I_2$$
$$\Rightarrow I_{2} = \dfrac{2.1}{3.5} = 0.6\ A $$

Current through $$20\ \Omega $$ and $$5\ \Omega $$ will be the same i.e. $$0.6A$$ as they are connected in series.

Hence, Option $$B$$ is correct

$$\textbf{Alternate solution:}$$
In a parallel combination of resistor $$R_1$$ and $$R_2$$, we know that current through $$R_2$$ is given by-
$$I_2=\dfrac{R_1}{R_1+R_2}I$$,
In this case, $$R_1= 6\Omega+4\Omega= 10\Omega$$ and $$R_2= 20\Omega+5\Omega=25\Omega$$
$$\Rightarrow I_2=\dfrac{10}{10+25}\times 2.1A = 0.6A$$
This is current through $$5\Omega$$ resistor.
Question 4
1 / -0

An electric iron draws a current of 15 A from a 220 V supply, What is the cost of using iron for 30 min everyday for 15 days if the cost of unit (1 unit =1 kWhr) is 2 rupees ?

A
Rs 49.5

B
Rs 60

C
Rs 40

D
Rs 10

Solution

As we know that power(P)=v$$\times$$I
so P=15$$\times$$220=3300 watt
cost per unit=2 rupees
therefore,$$\dfrac {3300\times2\times15\times30}{60\times1000}$$=49.5 rupees.
Question 5
1 / -0

What is the power produced by an appliance marked " $$240 V$$, $$2 A"?$$

A
$$8.3\ W$$

B
$$60\ W$$

C
$$120\ W$$

D
$$480\ W$$

Solution

Relation between power$$(P)$$, voltage$$(v)$$ and current$$(I)$$-
$$ P = VI$$
Given,
$$V = 240\ V$$
$$I = 2\ A$$
$$\Rightarrow P = 240 \times 2$$
$$\Rightarrow P = 480\ W$$
Question 6
1 / -0

In a metallic conductor, electric current is thought to be due to the movement of:

A
ions

B
amperes

C
electrons

D
protons

Solution

Current is due to flow of charged particles. Current flow can be due to movement of ions or electrons. In metals, however, the protons are bounded and hence electrons are the only contributors to current.
Question 7
1 / -0

A $$24 \ V$$ potential difference is applied across a parallel combination of four $$6 \ \Omega$$ resistor. The current in each resistor is:

A
$$1 \ A$$

B
$$4 \ A$$

C
$$16 \ A$$

D
$$36 \ A$$

Solution

Since the resistors are connected in parallel,the voltage across each resistor is the same ,i.e $$24 \ V$$.

Therefore, current in each resistor can now simply be calculated using the relation,

$$ I = \dfrac{V}{R} \implies I = \dfrac{24}{6.0} = 4.0 \ A $$
Question 8
1 / -0

Electric current is to be passed from one body to another. For this purpose the two bodies must be joined by

A
cotton thread

B
plastic string

C
copper wire

D
rubber band

Solution

Copper wire being a conductor can conduct .
Question 9
1 / -0

An electric bulb works on $$230\ V$$ line and draws $$0.1\ A$$ current. The resistance of the filament is:

A
$$230\ \Omega $$

B
$$2300\ \Omega $$

C
$$23\ \Omega $$

D
$$2.3\ \Omega $$

Solution

Voltage, $$V=230\ V$$
Current, $$I = 0.1\ A$$
According to Ohm's law,
$$ V = IR$$
$$\Rightarrow R = \dfrac{V}{I}$$

$$ = \dfrac{230}{0.1}$$
$$ \Rightarrow R = 2300\ \Omega$$
Question 10
1 / -0

Four identical wires each having resistance $$R$$ are connected as shown in the figure. Equivalent resistance between A and B is:

A
$$\dfrac{R}{4}$$

B
$$\dfrac{R}{2}$$

C
$$4R$$

D
$$R$$

Solution

Simplified circuit is drawn. It is the series combination of the four resistances.

$$ R_{AB} = 4R$$