Electricity Test

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Electricity Test - 60

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following I-V graph represents ohmic conductors?

A

B

C

D

Solution

Ohm's law $$I=\dfrac{V}{R}$$ is an equation of straight line. Hence I-V characteristics for ohmic conductors is also straight line and its slope gives, $$m=\dfrac{1}{resistance}$$ of the conductor.
Option A
Question 2
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$$n$$ resistors each of resistance $$R$$ first combine to give maximum effective resistance and then combine to given minimum effective resistance. The ratio of the maximum to minimum resistance is:

A
$$n$$

B
$$n^2$$

C
$$n^2-1$$

D
$$n^3$$

Solution

Maximum effective resistance will be when all the resistors are in series combination. So,
$$R_{eff_{max}}=n\ R$$
Minimum effective resistance will be when all the resistors are in parallel combination. So,
$$R_{eff_{min}}=R/n$$

$$\therefore \ \dfrac {R_{eff_{max}}}{R_{eff_{min}}}=\dfrac {n\ R}{R/ n}=n^2$$
Question 3
1 / -0

Ravi connected three bulbs with the cells and a switch as shown. When switch is moved to ON position.

A
The bulb X will glow first.

B
The bulb Y will glow first.

C
The bulbs Z and X will glow first.

D
All the bulbs will glow simultaneously.

Solution

All bulbs will glow simultaneously and instantly without any delay. It is because the current is setup in the circuit the moment the switch is closed to provide it a closed path to flow. The three bulbs are in series, so equal current flows through all of them.
Question 4
1 / -0

You have been provided with four $$400$$$$\ \Omega$$ resistors each. The number of ways in which these can be combined to have different equivalent resistances is

A
Seven different combinations and seven different equivalent resistances

B
Eight different combinations and seven different equivalent resistances

C
Nine different combinations and eight different equivalent resistances

D
Ten different combinations and nine different equivalent resistances.
Question 5
1 / -0

Two coils have a combined resistance of $$12 \Omega$$, when combined series and $$5/3 \Omega$$, when connected in parallel. Their respective resistances are:

A
$$6$$ and $$6$$ ohms

B
$$10$$ and $$2$$ ohms

C
$$5$$ and $$7$$ ohms

D
$$4$$ and $$8$$ ohms

Solution

Let us assume the two resistances to be $$R_1$$ and $$R_2$$.
Let resistances in series and parallel be $$R_s$$ and $$R_p$$ respectively.
Given:
$$R_{s}=12 \Omega=R_1+R_2$$
So,
$$R_1+R_2=12$$ . . . . (i)
$$R_2=12-R_1$$ . . . . (ii)

And,
$$R_{p}=\dfrac{5}{3}\Omega \Rightarrow \dfrac{1}{R_{p}}=\dfrac{3}{5}$$
So,
$$\dfrac{3}{5}=\dfrac{1}{R_1}+\dfrac{1}{R_2}$$

$$\Rightarrow \dfrac{3}{5}=\dfrac{R_1+R_2}{R_1R_2}$$ . . . (iii)

Using $$eq. (i)$$ and $$eq. (ii)$$ in $$eq. (iii)$$
$$\dfrac{3}{5}=\dfrac{12}{R_1(12-R_1)}$$

$$\Rightarrow 12R_1-R_1^2=20$$
$$\Rightarrow R_1^2-12R_1+20=0$$

This is a quadratic equation in $$R_1$$. Let's solve it by factorisation.
$$R_1^2-10R_1-2R_1+20=0$$
$$R_1(R_1-10)-2(R_1-10)=0$$
$$(R_1-10)(R_1-2)=0$$
$$\Rightarrow R_1-10=0$$ or $$R_1-2=0$$
$$\Rightarrow R_1=10$$ or $$R_1=2$$

Putting $$R_1=10$$ in $$eq.(i)$$, $$R_2=2$$
or
Putting $$R_1=2$$ in $$eq.(i)$$, $$R_2=10$$

So option B is correct.

Question 6

1 / -0

Match the Column I with Column II.

	Column I		Column II
(A)	Smaller the resistance greater the current in a circuit	(p)	If the same voltage is applied across a resistance
(B)	Greater or smaller the resistance the current is same	(q)	If the same current is passed
(C)	Greater the resistance, smaller the power	(r)	When resistances are connected in series
(D)	Greater the resistance, greater the power	(s)	When resistances are connected in parallel

A-r, B-p, C-q, D-s

A-p, B-r, C-q, D-s

A-r, B-p, C-s, D-q

A-s, B-r, C-p, D-q

Solution

In a parallel combination of resistances, the voltage drop across all the resistors is the same and therefore, the current across the resistances depends on the value of resistance. If the resistance is small, the current flowing through it will be more.

In a series combination of the resistances, the current through all the resistors is constant whereas the voltage differs for the resistors.

Similarly, if the same voltage is applied across all the resistors, the power is inversely proportional to the resistance. So, greater the resistance, smaller will be the power and if the current is the same in the circuit, the power will be greater for the higher resistance.

So, $$A\rightarrow s$$

$$B\rightarrow r$$

$$C\rightarrow p$$

$$D\rightarrow q$$

Question 7
1 / -0

A metal plate can be heated by

A
passing either a direct or alternating current through the plate.

B
placing in a time varying magnetic field.

C
placing in a space varying magnetic field, but does not vary with time.

D
both (a) and (b) are correct

Solution

When a metal plate is getting heated, it may be due to the passage of direct current, alternating current, or even an induced current passing through the plate. A time-varying magnetic field produces an induced current in the plate, and hence, both (a) and (b) are correct.
Question 8
1 / -0

In parallel combination of n cells, we obtain:

A
More voltage

B
More current

C
Less voltage

D
Less current

Solution

In parallel combination of n cells the net resistance of cells becomes $$\cfrac{1}{n}$$ times. Resistance reduces and current increases.
Question 9
1 / -0

A wire of resistance $$12\: ohm/meter\: $$is bent to form a complete circle of radius $$10cm$$. The resistance between its two diametrically opposite points, A and B as shown in the figure is?

A
$$3\Omega$$

B
$$6\pi \Omega$$

C
$$6\Omega$$

D
$$0.6\pi \Omega$$

Solution

Total length of wire, $$l= 2\pi\times 0.1=0.2\pi\ m$$
Resistance per unit length, $$r=12\Omega/m$$
Let us assume two semicircular part of wire connected in parallel between $$A$$ and $$B$$

Resistance of each part $$,R= r \times l = 12\times \cfrac{0.2\pi}{2}= 1.2 \pi\ \Omega$$

Two equal resistors (R) are connected in parallel between A and B so equivalent resistance is equal to $$R/2$$.

$$\therefore$$ Resistance between $$A$$ and $$B = \dfrac R2 = \dfrac{1.2 \pi}2= 0.6\pi \ \Omega$$

Hence, option $$(D)$$ is correct.
Question 10
1 / -0

An infinite ladder network of resistances is constructed with $$1\Omega$$ and $$2\Omega$$ resistance as shown in figure. The $$6$$V battery between A and B has negligible internal resistance. The equivalent resistance between A and B is?

A
$$1\Omega$$

B
$$2\Omega$$

C
$$3\Omega$$

D
$$4\Omega$$

Solution

$$ \textbf {Step 1: Find Smallest Repeating unit [Refer Fig. 1]} $$

$$ \textbf{Step 2: Replace with equivalent resistance R. [Refer Fig. 2]} $$
After the first smallest repeating unit, the remaining circuit is also same as the initial circuit.

Therefore, if we assume the equivalent resistance between A & B as $$R$$, then the resistance of the remaining circuit will also be $$R$$ as shown in the figure.

$$ \textbf {Step 3: Find equivalent resistance between A and B and equate with R. [Refer Fig. 3]} $$
From figure, the equivalent resistance between A & B will be equal to the equivalent resistance of the remaining circuit. So,

$$ R_{AB} = 1 + \dfrac {2R}{2+R} $$

$$\Rightarrow\ \ R = 1 + \dfrac{2R}{2+R} $$

$$\Rightarrow\ \ 3R + 2 = R^2 + 2R $$

$$\Rightarrow\ \ R^2 - R - 2 = 0 $$

$$\therefore\ \ \ \ \boxed{R = 2 \Omega}$$ (Here, Negative value can is rejected)

Hence Equivalent resistance between points A and B is 2$$\Omega$$.

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Electricity Test - 60

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Electricity Test - 60

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Question 1

Solution

Question 2

$$n$$

$$n^2$$

$$n^2-1$$

$$n^3$$

Solution

Question 3

The bulb X will glow first.

The bulb Y will glow first.

The bulbs Z and X will glow first.

All the bulbs will glow simultaneously.

Solution

Question 4

Seven different combinations and seven different equivalent resistances

Eight different combinations and seven different equivalent resistances

Nine different combinations and eight different equivalent resistances

Ten different combinations and nine different equivalent resistances.

Question 5

$$6$$ and $$6$$ ohms

$$10$$ and $$2$$ ohms

$$5$$ and $$7$$ ohms

$$4$$ and $$8$$ ohms

Solution

Question 6

A-r, B-p, C-q, D-s

A-p, B-r, C-q, D-s

A-r, B-p, C-s, D-q

A-s, B-r, C-p, D-q

Solution

Question 7

passing either a direct or alternating current through the plate.

placing in a time varying magnetic field.

placing in a space varying magnetic field, but does not vary with time.

both (a) and (b) are correct

Solution

Question 8

More voltage

More current

Less voltage

Less current

Solution

Question 9

$$3\Omega$$

$$6\pi \Omega$$

$$6\Omega$$

$$0.6\pi \Omega$$

Solution

Question 10

$$1\Omega$$

$$2\Omega$$

$$3\Omega$$

$$4\Omega$$

Solution

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