Answer the questions.[Phy-3M]

Question 513 Marks

Consider the circuit given below where A, B and C are three identical light bulbs of constant resistance.

List the bulbs in order of increasing brightness.
If C burns out, what will be the brightness of A now compared with before?
If B burns out instead, what will be the brightness of A and C compared with before?

Answer

C will be the brightest. Voltage will be distributed equally between A and B, so they will have equal brightness but lesser than that of C.
A gets the same voltage as before, so its brightness remains the same.
If B burns put, A will also stop glowing because it is connected in series with B. However, brightness of C remains the same.

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Question 523 Marks

Calculate the work done in moving a charge of 4 coulombs from a point at 220 volts to another point at 230 volts.

Answer

V₁ = 220 V, V₂ = 230V, Charge moved = 4C

Thus, the potential difference = V₂ - V₁ = 230 - 220 =10.

We know that,

Work done = Potential difference × Charge moved

= 10 × 4

Work done = 40 joules.

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Question 533 Marks

Calculate the area of cross-section of a wire if its length is 1.0m, its resistance is $23 Ω$ and the resistivity of the material of the wire is $1.84\times10 Ω\ \text{m}.$

Answer

I = 1.0m

R = 23 ohm

$\rho=1.84\times10^{-6}\ \text{ohm-meter}$

We have

$\text{R}=\rho\frac{\text{I}}{\text{A}}$

$23=1.84\times10^{-6}\times\frac{1}{\text{A}}$

$\text{A}=\frac{1.84\times10^{-6}}{23}$

$=0.08\times10^{-6}\text{m}^2$

$=8\times10^{-8}\text{m}^2$

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Question 543 Marks

A wire of given material having length l and area of cross-section A, has a resistance of $2\Omega$. Find the resistance of another wire of same material having length 2l and are of cross-section
$\frac{\text{A}}{2}$.

Answer

Let the resistivity of the material is s. now

Its length is given = l

Area of cross section = A

Resistan a $=4\Omega$

We know

$\text{R}=\text{s}\frac{1}{\text{a}}$

Now in case = ii

Resistance = s

Lenght = 21

Area of section$ = \frac{\text{A}}{2}$

So,

Resistance$={\text{p}}\frac {1}{\text{a}}$[p is constant]

For firest wire length = l and area of cross section = a

Resistance $= \frac{p}{a} = 4\text{ohm}$ [given]

For second wire lenght = 2L and area of cross section $= \frac{\text{A}}{2}$

Resistance $= \text{p } 2\text{L }\frac{\text{A}}{2}$

$= \text{P} \frac{\text{L}}{\text{A}}$

$= \text{P} \frac{\text{L}}{\text{A }} [\text{p }\frac{\text{L}}{\text{a}} = 4 \text{ohm}]$

$= 4 \text{ohm}$

The resistance of second wire = 4 ohm.

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Question 553 Marks

A student made an electric circuit as shown below.

Is there any mistake in this circuit? If any then correct it.

Answer

Here, in this circuit, the student has kept the ammeter in parallel and the voltmeter in series with the circuit.
But actually, the ammeter should always be connected in series as it has a very low resistance so when we connect the ammeter in series, maximum current will pass through it and thus the correct amount of current through the circuit can be measured.
In the same way, the voltmeter should always be connected in parallel as it is highly resistive due to which the current cannot pass through it and thus the current will pass through the component connected in parallel to it so as to measure the accurate voltage drop across the component.

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Question 563 Marks

A $4 Ω$ coil and a $2 Ω$ coil are connected in parallel. What is their combined resistance? A total current of 3A passes through the coils. What current passes through the $2 Ω$ coil?

Answer

$4\Omega$ and $2\Omega$ coil are connected in parallel.

Combined resistance is R

$\frac{1}{\text{R}}=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}$

$\text{R}=\frac{4}{3}\Omega$

Total current $\text{I}=\frac{\text{V}}{\text{R}}=3\Omega$

$\frac{\text{V}}{\frac{4}{3}}=3$

$\text{V}=3\times\frac{4}{3}=4\text{V}$

Current through $2\Omega$ coil $=\frac{\text{V}}{2}=\frac{4}{2}=2\text{A}$

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Question 573 Marks

A resistor has a resistance of 176 ohms. How many of these resistors should be connected in parallel so that their combination draws a current of 5 amperes from a 220 volt supply line?

Answer

I = 5A

V = 220V

$\text{R}=\frac{\text{V}}{\text{I}}=\frac{220}{5}=44\Omega$

Required resistance is less that $176\Omega,$ so the resistor should be connected in parallel.

Let the required no. be n.

$\text{R}_\text{eq}=\frac{176}{\text{n}}=44$

$\text{n}=\frac{176}{44}=4$

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Question 583 Marks

A p.d. of 6V is applied to two resistors of $3 Ω$ and $6 Ω$ connected in parallel. Calculate:
The combined resistance.
The current flowing in the main circuit.
The current flowing in the $3 Ω$ resistor.

Answer

V = 6V,

R₁ = 3 ohm, R₂ = 6 ohm (in parallel)

Combined resistance, $\frac{1}{\text{R}}=\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}$

$\frac{1}{\text{R}}=\frac{1}{3}+\frac{1}{6}=\frac{3}{6}=\frac{1}{2}$

R = 2 ohm

Current flowing in the main circuit, $\text{I}=\frac{\text{V}}{\text{R}}=\frac{6}{2}=3\text{A}$

Current flowing in 3 ohm resistor $=\frac{\text{V}}{\text{R}_1}=\frac{6}{3}=2\text{A}$

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Question 593 Marks

An electric bulb is rated as 10W, 220V. How many of these bulbs can be connected in parallel across the two wires of 220V supply line if the maximum current which can be drawn is 5A?

Answer

P = 10W, V = 220V, I = 5A
We know that
P = VI
= 220 X 5
P = 1100W
Power of one bulb = 10W
Total no. of bulbs that can be connected $=\frac{1100}{10}=110$

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Question 603 Marks

Aluminium wire has radius 0.25mm and length or 75m. If the resistance of the wire is $10\Omega$ calculate the resistivity of aluminium.

Answer

Here you are asked to find the resistivity of aluminium.

Given,

$\text{L}=75\text{cm}$

$\text{r}=25\times10^{-5}\text{m}$

$\text{R}=10\Omega$

$\text{P}= \ ?$

$\text{CSA}=\pi\text{r}^{2}$

$\text{A}=3.142\times(25\times10^{-5})^{2}\ \text{m}^{2}$

We know that

$\text{R}=\rho\frac{\text{L}}{\text{A}}$

$\Rightarrow\rho=\frac{\text{RA}}{\text{L}}$

$\Rightarrow\rho=10\times3.142\frac{(25\times10^{-5})}{75}$

$\Rightarrow\rho=2.6\times10^{-8}\Omega\text{m}$

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Question 613 Marks

You have four resistors of $8\Omega.$ each. Show how would you connect these resistors to have effective resistance of $8\Omega.$

Answer

Connect two resistors in series,the total resistance is R₁= 8 + 8 =16

then connect the other two resistors in series,the total resistance is R₂= 8 + 8 = 16

Then connect the two pairs in parallel.

that is,

let the final total be R

$\frac{1}{\text{R}}=\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}$

$\frac{1}{\text{R}}=\frac{1}{16}+\frac{1}{16}$

$\frac{1}{\text{R}}=\frac{2}{16}$

$\text{R}=8$

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Question 623 Marks

You are given three resistors each of $3\Omega$ and you are asked to get all possible values of resistance when you connect them in different combinations. How many values of resistance can you get?

Answer

Here, $\text{R}_{1}=3\Omega$

$\text{R}_{2}=3\Omega$

$\text{R}_{3}=3\Omega$

Parallel combination:

R₁, R₂ and R₃ connected in stair like pattern one above the other.

$\frac{1}{\text{R}_{\text{eq}}}=\frac{1}{\text{R}_{1}}+\frac{1}{\text{R}_{2}}+\frac{1}{\text{R}_{3}}$

$\frac{1}{\text{R}_{\text{eq}}}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}$

$\frac{1}{\text{R}_{\text{eq}}}=\frac{3}{3}$

$\frac{1}{\text{R}_{\text{eq}}}=\frac{3}{3}\Omega$

$\frac{1}{\text{R}_{\text{eq}}}=1\Omega$

Series Combination:

R₁, R₂ and R₃ connected in same line across the potential difference V

$\frac{1}{\text{R}_{\text{eq}}}=\text{R}_{1}+\text{R}_{2}+\text{R}_{3}$

$\frac{1}{\text{R}_{\text{eq}}}=3+3+3$

$\frac{1}{\text{R}_{\text{eq}}}=9\Omega$

Mixed combination:

R₁ is connected in series with the parallel combination fo R₂ and R₃.

$\frac{1}{\text{R}_{\text{q}}}=\frac{1}{\text{R}_{2}}+\frac{1}{\text{R}_{3}}$

Putting the values of the resistance, we get

$\frac{1}{\text{R}_{\text{q}}}=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}$

Thus, we get $\frac{1}{\text{R}_{\text{q}}}=1.5\Omega$

$\text{R}_{\text{s}}=\text{R}_{1}+\text{R}_{\text{p}}=3+1.4.5\Omega$

Thus, the net resistance of the circuit is $405\Omega$

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Question 633 Marks

Write the advantages of connecting electrical appliances in parallel and disadvantages of connecting them in series in a household circuit.

Answer

parallel connection:

Advandages:

Every unit that is connected in a parallel circuit gets equal amount of voltage.
It becomes easy to connect or disconnect a new element without affecting the working of other elements.
If any fault happened to the circuit, then also the current is able to pass through the circuit through different paths.

Series Connection:

We do not use series combination for connecting electrical appliances in household circuit as whenever there will be a damage/breakage in the circuit of any one appliance of the household then, due to the series connection all other connections.

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Question 643 Marks

With the help of a circuit diagram, obtain the relation for the equivalent resistance of two resistances connected in parallel.
In the circuit diagram shown below, find:

Total resistance.
Current shown by the ammeter A

Answer

Suppose total current flowing the circuit is I then the current passing through resistance R₁ will be I₁ and current passing through resistance R₂will be I₂

Total current = I = I₁ + I₂

Let resultant resistance of this parallel combination is R. By applying the ohm's law to each resistance we get that,

$\text{I}_1=\frac{\text{V}}{\text{R}_1}$

$\text{I}_2=\frac{\text{V}}{\text{R}_2}$

putting these eq in the above one, we get that

$\frac{\text{V}}{\text{R}}=\frac{\text{V}}{\text{R}_1}+\frac{\text{V}}{\text{R}_2}$

$\frac{1}{\text{R}}=\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}$

If two resistance are connected in parallel than the resultant resistance will be

$\frac{1}{\text{R}}=\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}$

Total resistance = R

$\frac{1}{\text{R}}=\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}$

R₂ = 3 + 2 = 5 ohms

R₁ = 5 ohm

$\frac{1}{\text{R}}=\frac{1}{5}+\frac{1}{5}$

$\frac{1}{\text{R}}=\frac{2}{5}$

R = 2.5 ohm

Current flows through the circuit

$\text{I}=\frac{\text{V}}{\text{R}}=\frac{4}{2.5}$

= 1.6 amps

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Question 653 Marks

What would be the effect on the resistance of a metal wire of:

Increasing its length?
Increasing its diameter?
Increasing its temperature?

Answer

Resistance will increase.
Resistance will decrease.
Resistance will increase.

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Question 663 Marks

What will be the length of a nichrome wire resistance $5.0\Omega$ if the length of similar wire of 120cm has resistance of $2.5\Omega$? Why?

Answer

Here,

$\text{I}_{2}=120\text{cm}=1.2\text{cm}$

$\text{R}_{1}5.0\Omega$

$\text{R}_{2}2.5\Omega$

$\therefore\frac{\text{R}_{1}}{\text{R}_{2}}=\frac{\rho\times\frac{\text{I}_{1}}{\text{A}}}{\rho\times\frac{\text{I}_{2}}{\text{A}}}$

$\frac{\text{R}_{1}}{\text{R}_{2}}=\frac{\text{I}_{1}}{\text{I}_{2}}=\frac{\text{I}_{1}}{120\times10^{-2}}$

$\frac{5.0}{2.5}=\frac{\text{I}_{1}}{120\times10^{-2}}$

$\text{I}_{1}2\times120\times10^{-2}\text{m}$

$\text{I}_{1}=24\times10^{-1}\text{m}$

$\therefore\text{I}_{1}=240\text{cm}$

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Question 673 Marks

What possible values of resultant resistance one can get by combining two resistances, one of value 2 ohm and the other 6 ohm?

Answer

R₁ = 2 ohm, R₂ = 6 ohm

Case I: (parallel combination)

$\frac{1}{\text{R}}=\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}$

$\frac{1}{\text{R}}=\frac{1}{2}+\frac{1}{6}=\frac{4}{6}$

$\text{R}=\frac{6}{4}=1.5\ \text{ohm}$

CaseII: (Series combination)

R = R₁+ R₂ = 2 + 6 = 8ohm

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Question 683 Marks

What is the difference between a good conductor and a poor conductor? Give two example of each.

Answer

	Good conductor	Poor conductor
1.	Good conductor have large numbers of free electrons.	Poorconductor have very few number of free electron.
2.	Good conductor have less resistance to the flow of electric current.	Poor conductor have high resistance to the flow of electric current.
3.	Good conductor: Copper and aluminium.	Poor conductor: wood and plastic.

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Question 693 Marks

What is Ohm’s law? Explain how it is used to define the unit of resistance.

Answer

Ohm’s law gives a relationship between current (I) and potential difference (V). According to ohm’s law: At constant temperature, the current flowing through a conductor is directly proportional to the potential difference across its ends. If I is the current flowing through a conductor and V is the p.d. across its ends, then according
to the ohm’s law:

$\text{I}\propto\text{V}$

or, $\text{I}\propto\text{V}$

or, V = RI

or, $\text{R}=\frac{\text{V}}{\text{I}}$

where, R is a constant called ''resistance'' of the conductor.

The unit of resistance is ohm.

If V = 1 volt and I = 1 amp, then $\text{R}=\frac{1}{1}=1\text{ohm}$

Thus, 1 ohm is the resistance of a conductor such that when a potential difference of 1 volt is applied to its ends, a current of 1 amp flows through it.

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Question 703 Marks

What is electrical resistivity? In a series electrical circuit comprising a resistor made up of a metallic wire, the ammeter reads 5A. The reading of the ammeter decreases to half when the length of the wire is doubled. Why?

Answer

The inherent property of a conductor because of which it resists the flow of electric current is called resistivity. Resistivity for a particular material is unique.
Resistance varies directly as length of the conductor.
Current varies inversely as resistance.
So, when length of the wire is doubled, its resistance becomes double. When resistance becomes double, current becomes half.
This explains why the reading of ammeter decreases to half when the length of the wire is doubled.

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Question 713 Marks

What is a circuit diagram? Draw the labelled diagram of an electric circuit comprising of a cell, a resistor, an ammeter, a voltmeter and a closed switch (or closed plug key). Which of the two has a large resistance: an ammeter or a voltmeter?

Answer

A diagram which indicates how different components in a circuit have been connected by using the electrical symbols for the components is called a circuit diagram.

A voltmeter has a large resistance.

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Question 723 Marks

What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

Answer

Advantages of connecting electrical devices in parallel:

When the appliances are connected in parallel with the battery, each appliance gets the same potential difference as that- of battery which is not possible in series connection.
Each appliance has different resistances and requires different currents to operate properly. This is possible only in parallel connection, as in series connection, same current flows through all devices, irrespective of their resistances.
If one appliance fails to work, other will continue to work properly.

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Question 733 Marks

Two resistances when connected in parallel give resultant value of 2 ohm; when connected in series the value becomes 9 ohm. Calculate the value of each resistance.

Answer

Two resistance when connected in series, resultant value is 9 ohms.

Two resistance when connected in parallel, resultant values is 2 ohms.

Let the two resistance be R₁ and R₂

If connected in series, then

9 = R₁+ R₂

R₁ = 9 - R₂

If connected in parallel, then

$\frac{1}{2}=\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}$

From aboves equations we get that

$\frac{1}{2}=\frac{(\text{R}_1+\text{R}_2)}{\text{R}_1\text{R}_2}$

$\frac{1}{2}=\frac{9}{(9-\text{R}_2)\text{R}_2}$

$9\text{R}_2-\text{R}_2{^2}=18$

$\text{R}_2{^2}-9\text{R}_2+18=0$

$(\text{R}_2-6)(\text{R}_2-3)=0$

$\text{R}_2=6,3$

So if R₂ 6 ohms, then R₁ = 9 - 6 = 3 ohms.

If R₂ = 3 ohms, then R₁ = 9 - 3 = 6 ohms.

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Question 743 Marks

Three 2V cells are connected in series and used as a battery in a circuit.

What is the p.d. at the terminals of the battery?
How many joules of electrical energy does 1C gain on passing through

One cell.
All three cells?

Answer

If three cells of 2 volt each are connected in series to make a battery, then the total potential difference between terminals of the battery will be 6V.

Given: p.d. = 2V, Charge moved = 1C

We know that

Work done = p.d × charge moved

= 2 × 1

Work done = 2 joules

Given: p.d = 6V, Charge moved = 1C

Work done = p.d × charge moved

= 6 × 1

Work done = 6 joule.

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Question 753 Marks

The electrical resistivities of three materials P, Q and R are given below:

$\begin{matrix}\text{P}&2.3\times10^{3}\Omega\text{ m}\\\text{Q}&2.63\times10^{-8}\Omega\text{ m}\\\text{R}&1.0\times10^{15}\Omega\text{ m}\end{matrix}$

Which material will you use for making:

Electric wires.
Handle for soldering iron, and
Solar cells?

Give reasons for your choices.

Answer

Material Q with resistivity 2.63 × 10-8 ohm-m can be used for making electric wires because it has very low resistivity.
Material R with resistivity 1.0 × 1015 ohm-m can be used for making handle of soldering iron because it has very high resistivity.
Material P with resistivity 2.3 × 103 ohm-m can be used for making solar cell because it is a semiconductor.

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Question 763 Marks

The electrical resistivities of four materials P, Q, R and S are given below:

$\begin{matrix}\text{P}&6.84\times10^{-8}\Omega\text{m}\\\text{Q}&1.70\times10^{-8}\Omega\text{m}\\\text{R}&1.0\times10^{15}\Omega\text{m}\\\text{S}&11.0\times10^{-7}\Omega\text{m}\end{matrix}$

Which material will you use for making:

Heating element of electric iron.
Connecting wires of electric iron.
Covering of connecting wires?

Give reason for your choice in each case.

Answer

S; because it has high resistivity of $\frac{11}{10000000}\text{ohm\ m}$ (it is actually nichrome).
Q; because it has very low resistivity of $\frac{1.7}{100000000}\text{ohm\ m}$ (it is actually copper).
R; because it has very very high resistivity of$1.0\times100000000000000\ \text{ohm\ m}$ (it is actually rubber).

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Question 773 Marks

Ten bulbs are connected in a series circuit to a power supply line. Ten identical bulbs are connected in a Parallel circuit to an identical power supply line.

Which circuit would have the highest voltage across each bulb?
In which circuit would the bulbs be brighter?
In which circuit, if one bulb blows out, all others will stop glowing
Which circuit would have less current in it?

Answer

Parallel Circuit

because all will have same voltage across them i.e. voltage of battery.

Parallel Circuit

because all have higher voltage across them.

Series Circuit

because if one bulb goes out current won't reach the next bulb.

Series Circuit

because Equivalent Resistance will be less and more current will be drawn in parallel circuit.

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Question 783 Marks

State and explain Joule’s law of heating.

Answer

Joule's law of heating states that heat produced in joule when a current of I amphere flows in a wire of resistance R ohms for time t secounds is
Given by H = I²Rt
Thus the heat produced in a wire is directly proportional to:

Square of current.
Resistance of wire.
Time for which current is passed.

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Question 793 Marks

In which of the following cases more electrical energy is consumed per hour?

A current of 1 ampere passed through a resistance of 300 ohms.
A current of 2 amperes passed through a resistance of 100 ohms.

Answer

R = 300 ohm, I = 1A, t = 1h

P = I² R = 1² × 300 = 300W

E = P × t = 300 × 1 = 300Wh

R = 100 ohm, I = 2A, t = 1h

P = I² R = 2² × 100 = 400W

E = P × t = 400 × 1 = 400Wh

Hence, in case (ii), the electrical energy consumed per hour is more.

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Question 803 Marks

In the circuit diagram given below, the current flowing across 5 ohm resistor is 1 amp. Find the current flowing through the other two resistors.

Answer

Given:
1 amp current is flowing through 5ohm resistor.
We know that in case of parallel connection, the p.d. across each resistor is same and is equal to the voltage applied.
Therefore, applied voltage, V = IR = 1 × 5 = 5V
So,
Current through 4 ohm resistor = $\frac{\text{V}}{\text{R}}=\frac{5}{4}=1.25\text{A}$
Current through 10 ohm resistor = $\frac{\text{V}}{\text{R}}=\frac{5}{10}=0.5\text{A}$

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Question 813 Marks

In a household electric circuit, different appliances are connected in parallel to one another. Give two reasons. An electrician puts a fuse of rating 5A in that part of domestic electrical
circuit in which an electrical heater of rating 1.5kW, 220V is operating. What is likely to happen in this case and why? What change, if any, needs to be made?

Answer

Each appliance will be at the same potential i.e., voltage.
If one of the appliances fail, the others will still keep working. Rating of fuse = 5A,

V = 220V,

P = 1.5 kW = 1500W

Current in the circuit

$\text{I} = \frac{\text{P}}{\text{V}} = \frac{1500}{220} = 6.8\text{A }$

The current in the circuit is more than the rating of the fuse. Therefore, the fuse will blow off. A fuse of rating of about 10A should be put in the circuit.

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Question 823 Marks

If the length of a wire is halved and its cross-sectional area is doubled, then what would be the resistance of the wire? (Given, initially the resistance of the wire is R)

Answer

The formula of resistance is: $\text{R}=\frac{\rho\times\text{I}}{\text{A}}$

Where, $\rho=$ resistivity

I = Length of the conductor

A = Area of the conductor

If length of the conductor is halved then, $\text{I}_{1}=\frac{1}{2}$

And, the cross sectional area is doubled, $\text{A}_{1}=2\text{A}$

$\therefore\text{R}_{1}=\frac{\rho\times\text{I}_{1}}{\text{A}_{1}}$

$\text{R}_{1}=\frac{\rho\times\text{I}_{1}}{\text{A}_{1}}$

$\text{R}_{1}=\frac{\rho\times\frac{1}{2}}{\text{A}_{1}}$

$\text{R}_{1}=\frac{1}{4}\times\frac{\rho\times\text{I}}{\text{A}}$

Now, $\frac{\rho\times}{\text{A}}=\text{R}$

$\therefore\text{R}_{1}=\frac{\text{R}}{4}$

Thus, we can conclude that the new resistance of the wire when the area of wire is doublead and length is haved become the initial resistance.

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Question 833 Marks

How much energy is consumed when a current of 5 amperes flows through the filament (or element) of a heater having resistance of 100 ohms for two hours? Express it in joules.

Answer

I = 5 amp, R = 100 ohms, t = 2h
We know that
Electric energy consumed = P x t = I * I * Rt
= 25 × 100 × 2
= 5000Wh
= 5kwh
We know that 1kwh = 3.6 × 106J
Therefore, 5kwh = 5 × 3.6 × 106J = 18 × 106J.

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Question 843 Marks

How does the resistance of a wire change when:

Its length is tripled?
Its diameter is tripled?
Its material is changed to one whose resistivity is three times?

Answer

$\text{R}=\rho\frac{\text{I}}{\text{A}}$

I → 3I

$\text{R}'=\rho\frac{3\text{I}}{\text{A}}=3\text{R}$

Resistance gets tripled.

d → 3d

$\text{R}=\rho\frac{\text{I}}{\text{A}}=\text{R}=\rho\frac{\text{I}}{\pi\text{r}^2}=\rho\frac{\text{I}}{\pi\big(\frac{\text{d}}{2}\big)^2}$

$\text{R}'=\rho\frac{\text{I}}{\pi\big(\frac{3\text{d}}{2}\big)^2}=\frac{1}{9}\rho\frac{\text{I}}{\pi\big(\frac{\text{d}}{2}\big)^2}=\frac{\text{R}}9{}$

Resistance becomes $\frac{1}{9}\text{th}.$

$\text{R}=\rho\frac{\text{I}}{\text{A}}$

$\rho\rightarrow3\rho$

$\text{R}'=3\rho\frac{\text{I}}{\text{A}}=3\text{R}$

Resistance becomes 3 times.

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Question 853 Marks

How does the resistance of a conductor depend on:

Length of the conductor?
Area of cross-section of the conductor?
Temperature of the conductor?

Answer

Resistance of a conductor increases (or decreases) with increase (or decrease) in the length of the conductor.
Resistance of a conductor decreases (increases) with increase (decrease) in the area of cross-section of the conductor.
Resistance of a conductor increases on raising the temperature and decreases on lowering the temperature.

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Question 863 Marks

For the circuit shown in the diagram below:

What is the value of :
Current through $6 Ω$ resistor?
Potential difference across $12 Ω$ resistor?

Answer

As per the circuit
V = 4V
Total resistance in line 1 = R₁ = 6 + 3 = 3 ohm
Total resistance in line 2 = R₂ = 12 + 3 = 15 ohm
Current through $6\Omega$ resistor = current through line $1=\frac{\text{=V}}{\text{R}_1}=\frac{4}{9}=0.44\Omega$
p.d. across line 2 is 4V
current through line $2=\frac{\text{V}}{\text{R}_2}=\frac{4}{15}\Omega$
p.d. across $12\Omega$ resistor $=\frac{4}{15}\times12=3.2\text{V}$

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Question 873 Marks

Find the equivalent resistance across the two ends A and B of the following circuits.

Answer

Pand are of the some poteotiol means there no currunt flows be tween p and q
So,

$\frac{1}{\text{R}_\text{T}}=\frac{1}{\text{R}}+\frac{\text{1}}{\text{R}}$
$\text{R}_\text{T}= \frac{\text{R}}{2}$

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Question 883 Marks

Find the equivalent resistance across the two ends A and B of the following circuit.

Answer

R₁ and R₂ are in parallel

So,

$\frac{1}{\text{R}^1}=\frac{1}{\text{R}_2}+\frac{1}{\text{R}_2}$

$\frac{1}{\text{R}^1}=\frac{1}{2}+\frac{1}{2}$

$\frac{1}{\text{R}_1}=\frac{2}{2}=1$

$\text{R}^1=1\text{ohm.}$

R'' (R₃ and R₄ are also in parallel ) = 1ohm.

Now, R¹ and R'' are in series

so,

R^||| = R^| + R^||

= 1 + 1

= 2 ohm

Now,

R^||| (R₁,R₂,R₃,R₄) and R^|||| (R₅,R₆,R₇,R₈)

Are in the parallel,

(Total resistance)

$\frac{1}{\text{R}}=\frac{1}{\text{R}^{|||}}+\frac{1}{\text{R}^{|||}}$

$=\frac{1}{2}+\frac{1}{2}$

$=\frac{1}{\text{R}}+\frac{2}{2}=1$

$\text{R}=1\text{ohm}$

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Question 893 Marks

Electrical resistivities of some substances, in ohmere, at 20°C are given as follows:

Silver	1.60 x 10^-8
Copper	1.62 x 10^-8
Tungsten	5.2 x 10^-8
Mercury	94 x 10^-8
Iron	10 x 10^-8
Nichrome	10 x 10^-6

Out of the two silver and copper, which one is better conductor of electric current and why?
Which substance is preferred to be used for electrical transmission lines? Give reason.
Name the material that you would advice to use in the heater element of electric heating device and why?

Answer

Silver is a better conductor, lower resistivity.
Copper, economical/low resistivity.
Nichrome; Very high resistivity/ as it is an alloy, it does not oxidize readily at high temperature.

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Question 903 Marks

Distinguish between resistance and resistivity.

Answer

Resistance is the property of the conductor, while resistivity is the property of the material of the conductor.
Resistance of a conductor is the opposition to the flow of electric current through it. resistivity of a substance is the opposition to the flow of electric current by a rod of that substance which is 1m long and 1m² in cross section.
Resistivity of a substances depends on the nature of the substance and temperature.
Resistivity of a substance on the nature of the substance and its temper. it does not depend the length or thickness of the conductor.

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Question 913 Marks

Distinguish between good conductors, resistors and insulators. Name two good conductors, two resistors and two insulators.

Answer

Those substances which have very low electrical resistance are called as good conductors. E.g., copper and aluminium.
Those substances which have comparatively high resistance than conductors are known as resistors. E.g., nichrome and manganin.Those substances which have infinitely high electrical resistance are called insulators. E.g., rubber and wood.

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Question 923 Marks

Derive the expression for the heat produced due to a current ‘I’ flowing for a time interval ‘t’ through a resistor ‘R’ having a potential difference ‘V’ across its ends. With which name is this relation known?

Answer

when an electric charge Q moves against a.p.d.V, the amount of work done is given by
W = Q × V .......(1)
We know, current, $\text{I}=\frac{\text{I}}{\text{T}}$
Q = I × t .....(2)
By ohm's law, $\frac{\text{V}}{\text{I}}=\text{R}$
V = I × R ........(3)
Putting eqs. (2) and (3) in eq (1),
W = I × t × I × R
W = I²RT
Assuming that all the electrical work done is converted into heat energy, we get Heat produced, H = I²Rt joules
This relation is known as Joule's law of heating,

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Question 933 Marks

Calculate the resistance of an aluminium cable of length 10km and diameter 2.0mm if the resistivity of aluminium is $2.7\times10 Ω\ \text{m}.$

Answer

I = 10km = 10000m

d = 2mm

r = 1mm = 10^-3m

$\rho=2.7\times10^{-8}\Omega\ \text{m}$

$\text{R}=\rho\frac{\text{I}}{\text{A}}$

$=2.7\times10^{-8}\times\frac{10000}{3.14\times(10^{-3})^2}$

$=0.859\times10^2\Omega$

$=86\Omega$

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Question 943 Marks

Calculate the resistance of a copper wire 1.0km long and 0.50mm diameter if the resistivity of copper is $1.7\times10\Omega\ \text{m}.$

Answer

I = 1km = 1000m

$\text{r}=\frac{\text{d}}{2}=\frac{0.5}{2}\text{mm}=0.25\text{mm}=0.25\times10^{-3}\text{m}$

$\rho=1.7\times10^{-8}\Omega\text{m}$

$\text{R}=\rho\frac{\text{I}}{\text{A}}=\rho\frac{\text{l}}{\pi\text{r}^2}$

$\text{R}=1.7\times10^{-8}\times\frac{1000}{3.14\times(0.25\times10^{-3})^2}=86.6\Omega$

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Question 953 Marks

Calculate the power used in the $2Ω$ resistor in each of the following circuits:

A 6V battery in series with $1Ω$ and $2Ω$ resistors.
A 4V battery in parallel with $12Ω$ and $2Ω$ resistors.

Answer

V = 6 Volt, R1 = $\text{R}_1=1\Omega,$ $\text{R}_2=2\Omega$

Equivalent resistance $=\text{R}_1+\text{R}_2=1+2=3\Omega$

Total current, $\text{I}=\frac{\text{V}}{\text{R}}=\frac{6}{3}=2\text{A}$

Current through R₂ = I₂= I = 2A

Voltage across R₂ = V₂ = I₂R₂ = 2 × 2 = 4

Power used in R₂ = I₂V₂= 2 × 4 = 8W

V = 4Volt, $\text{R}_1=12\Omega$ $\text{R}_2=2\Omega$

Voltage across R₂ = V₂ = V = 4V

Current across $\text{R}_2=\text{I}_2=\frac{\text{V}_2}{\text{R}_2}=\frac{4}{2}=2\text{A}$

Poower used in R₂ = I₂V₂ = 2 × 4 = 8W

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Question 963 Marks

Calculate the cost of operating a heater of 500W for 20hours at the rate of? ₹3.90 per unit.

Answer

Given P = 500W = 0.5KW, t = 20hr
We know that
Energy consumed = P × t = 0.5 × 20
= 10KWh
Total cost = 10 × cost per unit
cost per unit = Rs. 3.9 per unit
Therefore, total cost = 10 × 3.9 = Rs. 39

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Question 973 Marks

Bulb is rated at 200V, 100W. Calculate its resistance. Five such bulbs are lighted for 4 hours daily. Calculate the units of electrical energy consumed per day. What would be the cost of
using these bulbs per day at the rate of Rs. 4.00 per unit?

Answer

Using,$\text{P} = \frac{\text{V} 2 }{\text{R}}$

we get $\text{R} = \frac{\text{V} 2}{ \text{P}}$

$= 200\frac{2}{100}$

$= 400 $

Energy consumed = P × t

= 100 × 4

= 400 Wh

= 0.4kWh

Cost at the rate of 50 paise per unit = Rs (4 × 0.4)

= Rs 1.6

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Question 983 Marks

A wire is 1.0m long, 0.2mm in diameter and has a resistance of $10Ω.$ Calculate the resistivity of its material?

Answer

l = 1m

$\text{r}=\frac{\text{d}}{2}=\frac{0.2}{2}\text{mm}=0.1\text{mm}=0.0001\text{m}$

R = 10 ohm

We know that,

$\text{R}=\text{P}\frac{\text{I}}{\text{A}}$

$\text{P}=\frac{\text{RA}}{\text{I}}$

$=\frac{10\times\pi\times(0.0001)^2}{1}$

$=31.4\times10^{-8}\Omega\text{m}$

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Question 993 Marks

A resistor of 8 ohms is connected in parallel with another resistor X. The resultant resistance of the combination is 4.8 ohms. What is the value of the resistor X?

Answer

Given:

A resistor of 8ohm is connected in parallel with a resistor of x.

And resultant is 4.8.

Then X = ?

We know that for parallel case

$\frac{1}{\text{R}}=\frac{1}{\text{R}_1}+\frac{1}{\text{X}}$

$\frac{1}{4.8}=\frac{1.8}{}+\frac{1}{\text{x}}$

$\frac{1}{4.8}-\frac{1}{8}=\frac{1}{\text{x}}$

After solving we get that

X = 12 ohms.

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Question 1003 Marks

A resistance of 40 ohms and one of 60 ohms are arranged in series across 220 volt supply. Find the heat in joules produced by this combination of resistances in half a minute.

Answer

Given: R1 = 40 ohms, R2 = 60 ohms (in series), V = 220V, t = 30sec

we know that

Total resistance, R = 40 + 60 = 100 ohms

By ohm's law

V = IR

$\text{I}=\frac{\text{V}}{\text{R}}$

$\text{I}=\frac{220}{100}=2.2\text{amp}$

Putting the value of I, R and t in eq. H = I²RT

H = 2.2² × 100 × 30

H = 14520J.

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Science Answer the questions.[Phy-3M]