Electricity Test

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Electricity Test - 64

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Weekly Quiz Competition

Question 1
1 / -0

The current in the given circuit will be :

A
$$\dfrac{1}{45}A$$

B
$$\dfrac{1}{15}A $$

C
$$\dfrac{1}{10}A $$

D
$$\dfrac{1}{5}A $$

Solution

Two 30 ohm are in series equivalent resistance is 60 ohm
This 60 ohm and 30 ohm are in parallel its equivalent resistance is ,
$${ R }_{ eq }=\frac { 30\times 60 }{ 30+60 } =\frac { 1800 }{ 90 } =20\Omega $$
$$i=\frac { V }{ { R }_{ eq } } =\frac { 2 }{ 20 } =\frac { 1 }{ 10 } \Omega $$
Hence, option (C) is correct option
Question 2
1 / -0

Two resistors $$R_1 \, (24 \pm 0.5) \Omega$$ and $$R_2 \, (8 \pm 0.3) \Omega$$ are joined in series. The equivalent resistance is

A
$$32 \pm 0.33 \Omega$$

B
$$32 \pm 0.8 \Omega$$

C
$$32 \pm 0.2\Omega$$

D
$$32 \pm 0.5 \Omega$$

Solution

R₁ = 24 ± 0.5

R₂ = 8 ± 0.3

In series R₁+ R₂ = 32 ± 0.8
Question 3
1 / -0

In the given figure below, the current passing through $$6\Omega$$ resistor is

A
$$0.40A$$

B
$$0.48A$$

C
$$0.72A$$

D
$$0.80A$$

Solution

$$P.d$$ across the circuit $$=1.2\times \dfrac {6\times 4}{6+4}=2.88$$ volt

Current through $$6$$ ohm resistance $$=\dfrac {2.88}{6}=0.48\ A$$
Question 4
1 / -0

Two resistors $$R_1$$ and $$R_2$$ are connected in parallel. The relative error in their equivalent resistance is [If $$R_1=(60\pm 3)\Omega $$ and $$R_2=(30\pm 2)\Omega $$

A
$$\dfrac{1}{20}$$

B
$$\dfrac{1}{30}$$

C
$$\dfrac{11}{180}$$

D
$$\dfrac{7}{180}$$

Solution
Question 5
1 / -0

Three resistances $$R_1,R_2$$ & $$R_3$$ $$(R_1> R_2> R_3)$$ are connected in series. If current $$I_1,I_2$$ & $$I_3$$ respectively is flowing through them, the correct relation will be

A
$$I_1=I_2=I_3$$

B
$$I_1> I_2> I_3$$

C
$$I_1< I_2< I_3$$

D
$$I_1> I_2< I_3$$

Solution

In series, same amount of current will flow through all the resistances.
Current $$I_1=I_2=I_3=I=\dfrac{V}{R_{eq}}=\dfrac{V}{R_1+R_2+R_3}$$

Option A is correct.
Question 6
1 / -0

When three resistors of resistances $$3\Omega ,4\Omega $$ and $$5\Omega $$ are connected in parallel, the currents through them are in the ratio

A
$$5:4:3$$

B
$$20:15:12$$

C
$$3:4:5$$

D
$$9:8:7$$

Solution

Given:
Resistance $$R_{1} = 3\Omega$$
Resistance $$R_{2} = 4\Omega$$
Resistance $$R_{3} = 5\Omega$$
$$R_{1} / R_{2} / R_{3}$$
Since the resistors are parallel, voltage across them is same $$-V$$
Let $$I_{1}, I_{2}$$ and $$I_{3}$$ be current flowing through $$R_{1}|R_{2}, R_{3}$$ respectively.
Then; $$I_{1} = \dfrac {V}{3} A; I_{3} = \dfrac {V}{4}A; I_{3} = \dfrac {V}{5}A$$
$$I_{1} : I_{2} : I_{3} = \dfrac {V}{3} : \dfrac {V}{4} : \dfrac {V}{5} = \dfrac {1}{3} : \dfrac {1}{4} : \dfrac {1}{5}$$
i.e. $$I_{1} : I_{2} : I_{3} = 60 \left (\dfrac {1}{3} : \dfrac {1}{4} : \dfrac {1}{5}\right ) = 20 : 15 : 12$$
$$\therefore$$ Option B is correct.
Question 7
1 / -0

The value equlvalent resistance between the points $$A$$ and $$B$$ in the given circuit will be?

A
$$6\ R$$

B
$$\dfrac{4R}{11}$$

C
$$\dfrac{11R}{4}$$

D
$$\dfrac{R}{6}$$

Solution
Question 8
1 / -0

A uniform wire of resistance $$36\ \Omega$$ is bent in the form of a circle. The effective resistance across points A and B is___.

A
$$5 \Omega$$

B
$$6 \Omega$$

C
$$7.2 \Omega$$

D
$$30 \Omega$$

Solution
Question 9
1 / -0

The lowest resistance which can be obtained by connecting $$10$$ resistors each of $$\dfrac{1}{10}\Omega$$ is

A
$$1/250\ \Omega$$

B
$$1/100\ \Omega$$

C
$$1/200\ \Omega$$

D
$$1/10\ \Omega$$

Solution

The lowest resistance will be in the case when all the resistors are connected in parallel.
In parallel combination, we calculate the equivalent resistance $$(R_{eq})$$ as

$$\dfrac{1}{R_{eq}}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}} +\dfrac{1}{R_{3}}.........$$
Where $$R_1, R_2, R_3..... $$ are the resistances that we are connecting in the parallel.
In our question, All the 10 resistors have a resistance of $$\dfrac{1}{10}\Omega = 0.1\Omega$$

$$\dfrac{1}{R_{eq}}=\dfrac{1}{0.1}+\dfrac{1}{0.1}.........10\ times $$

$$\dfrac{1}{R_{eq}}=10+10......10\ times $$

$$\dfrac{1}{R_{eq}}=100$$

$$R_{eq}=\dfrac{1}{100}\Omega$$
Question 10
1 / -0

An electric bulb is rated $$220\ V$$ and $$100\ W$$, when it is operated on $$110\ V$$, the power consumed will be

A
$$100\ W$$

B
$$75\ W$$

C
$$50\ W$$

D
$$25\ W$$

Solution

D. $$25W$$

Formulae,

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

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

$$=\dfrac{220^2}{100}=484\Omega $$

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

$$P=\dfrac{110^2}{484}=25W$$