Electricity Test

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Electricity Test - 68

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Weekly Quiz Competition

Question 1
1 / -0

For the circuit shown in figure given below, the equivalent resistance between points A and B is

A
$$10\Omega$$

B
$$5\Omega$$

C
$$\dfrac{10}{3} \Omega$$

D
$$2\Omega$$

Solution
Question 2
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If $$5$$ resistors of $$2.5 \Omega$$ each are connected in parallel then what would be its resultant resistance?

A
$$5\Omega$$

B
$$0.5\Omega$$

C
$$50\Omega$$

D
$$6.25\Omega$$

Solution

Given:
Five resistances of $$R=2.5 \ \Omega$$ each are connected in parallel.
By formula of parallel combination,
$$\dfrac{1}{R_{eq}}=\dfrac{1}{R}+\dfrac{1}{R}+\dfrac{1}{R}+\dfrac{1}{R}+\dfrac{1}{R}$$
$$\Rightarrow \dfrac{1}{R_{eq}}=5 \times \dfrac{1}{R}=5 \times \dfrac{1}{2.5}$$
$$\Rightarrow R_{eq}=\dfrac{2.5}{5}=0.5 \ \Omega $$
So the resultant resistance is $$0.5 \ \Omega.$$
Question 3
1 / -0

If five equal pieces of $$25\ \Omega$$ are connected in parallel, then their equivalent resistance will be _____

A
$$25\Omega$$

B
$$1\Omega$$

C
$$5\Omega$$

D
$$\cfrac{1}{5}\Omega$$

Solution

In the parallel combination-

$$\dfrac{1}{R_{eq}} = \dfrac{1}{R_1} + \dfrac{1}{R_2}+\dfrac{1}{R_3}+\dfrac{1}{R_4} + \dfrac{1}{R_5}$$

Here all resistors are equal,
$$R_{1}=R_2=R_3=R_4=R_5=R=25\ \Omega$$

So,
$$\Rightarrow \dfrac{1}{R_{eq}} = \dfrac{5}{R}$$

$$\Rightarrow \dfrac{1}{R_{eq}} = \dfrac{5}{25}$$

$$\Rightarrow \dfrac{1}{R_{eq}} = \dfrac{1}{5}$$

$$\Rightarrow R_{eq} = 5\ \Omega$$
Question 4
1 / -0

A wire of cross-sectional area $$5.0\times 10^{-6}m^2$$ is made of a metal of resistivity $$50\times 10^{-8}\Omega$$m. The potential difference across the wire is $$6.0$$V and the current is $$3.0$$A. What is the length of the wire?

A
$$0.050$$m

B
$$0.20$$m

C
$$5.0$$m

D
$$20$$m

Solution

Given that,
Cross section area of wire , $$A= 5.0\times 10^{-6} \ m^2$$
Resistivity of wire , $$\rho= 50\times 10^{-8}\ \Omega m$$
Let $$l$$ be the length of wire
Potential difference across wire , $$V= 6.0 \ V$$
Current through the wire , $$I= 3.0 \ A$$

$$\therefore$$ Resistance of wire , $$R= \dfrac VI=\dfrac 63=2 \ \Omega$$

Also , $$R=\rho \dfrac l A$$

$$\implies 2=50\times 10^{-8} \times \dfrac l{5.0\times 10^{-6} }$$

$$\implies l= \dfrac{2\times 5.0\times 10^{-6}}{50\times 10^{-8}}=20\ m $$

$$\implies \text{Length of wire, }l =20\ m$$
Question 5
1 / -0

Three resistances are arranged as shown. The potential of the point O is :

A
3 V

B
1 V

C
4 V

D
5 V

Solution
Question 6
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Which unit is not used in either the definition of the coulomb or the definition of the volt?

A
ampere

B
joule

C
ohm

D
second

Solution

Coulomb is the unit of electric charge and volt is the unit of potential difference.
Ohm is the measure of resistance. It has nothing to do with the definition of volt or coulomb.
Question 7
1 / -0

A $$220 \ V$$ and $$100 \ W$$ lamp is connected to $$220 \ V$$ power supply. What is the power consumed?

A
$$100$$W

B
$$200$$W

C
More than $$100$$W

D
None of these

Solution

The power consumed would be obviously $$100 \ W$$.
It is because the lamp is operated at its rated voltage $$V =220 \ V$$. So the power consumed will be equal to its power rating, i.e. $$100 \ W.$$

Let us solve it to prove it.

We know, power $$P=\dfrac{V^2}{R}$$
Putting $$P=100 \ W$$ and $$V =220 \ V $$
$$\Rightarrow R=\dfrac{(220)^2}{100}=484 \ \Omega$$

By Ohm's law,
Current in the circuit $$I=\dfrac{\text{power supply voltage V}}{\text{resistance R}}$$
$$\Rightarrow I=\dfrac{220}{484} \ A$$

Power consumed, P can be calculated as $$I^2R.$$
$$P={\left( \dfrac{220}{484}\right)}^2 \times 484=\dfrac{(220)^2}{484}=100 \ W$$

Hence, option (A) is correct.
Question 8
1 / -0

A wire has resistance $$12\Omega$$. It is bent in the form of a circle. The effective resistance between the two points on any diameter of the circle is?

A
$$12\Omega$$

B
$$24\Omega$$

C
$$6\Omega$$

D
$$3\Omega$$

Solution
Question 9
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Three $$2\Omega$$ resistors are connected to form a triangle. The resistance between any two corners is?

A
$$6\Omega$$

B
$$2\Omega$$

C
$$\dfrac{3}{4}\Omega$$

D
$$\dfrac{4}{3}\Omega$$

Solution
Question 10
1 / -0

In the circuit shown, the current in $$3\Omega$$ resistance is?

A
$$1$$A

B
$$1/7$$A

C
$$5/7$$A

D
$$15/7$$A

Solution

We know,
The voltage of ground is considered to be 0V,

Let the grounds be named as 1,2,3 respectively,
And let the current in $$1^{st}$$ $$2\Omega$$ be i, and the $$3\Omega$$ be $$i_1$$,

Applying Kirchhoff Voltage law , from ground 1 to ground 2,

$$0+10-2i-2(i-i_1)=0$$
$$2i-i_1=5$$ $$-(i)$$

Applying KVL from ground 1 to ground 3,

$$0+10-2i-3i_1-3i_1=0$$
$$2i+6i_1=10$$ $$-(ii)$$

Subtracting $$(i)$$ from $$(ii)$$,

$$7i_1=5$$

$$i_1=\dfrac{5}{7} A$$

Option $$\textbf C$$ is the correct answer.