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Science Answer the questions.[Phy-3M]

Electricity · CBSE STD 10 (CBSE - English Medium). 62 questions.

Questions · Page 1 of 2

Answer the questions.[Phy-3M]

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Question 13 Marks
You are given one hundred $1 Ω $ resister. What is the smallest and largest resistance you can make in a circuit using these?
Answer
Given: n = 100, R = 1 ohm
For obtaining the smallest resistance, these resistance are connected in parallel:
Equivalent resistance $=\frac{1}{1}+\frac{1}{1}+\frac{1}{1}....100\ \text{times}=\frac{100}{1}$
Req $=\frac{1}{100}=0.01\ \text{ohm}$
For obtaining the largest resistance, these are connected in series:
Equivalent reisistance = 1 + 1 + 1 .........100 Times = 100
Req = 100 ohm
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Question 23 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:
  1. Total resistance.
  2. Current shown by the ammeter A

Answer


  1. Suppose total current flowing the circuit is I then the current passing through resistance R1 will be I1 and current passing through resistance R2 will 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

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

  1. 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 33 Marks
Which uses more energy: a 250W TV set in 1 hour or a 1200W to aster in 10 minutes?
Answer
Case1: TV set
P = 250W = 0.25 KWh
t = 1h
Energy consumed = P × t = 0.25 × 1 = 0.25KWh
Case2: Toaster
P = 1200W = 1.2KW, t = 10min $=\frac{10}{60}=\frac{1}{6}\text{h}$
Energy consumed $=\text{P}\times\text{t}=1.2\times\big(\frac{1}{6}\big)=0.2\text{kWh}$
Thus, TV uses energy.
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Question 43 Marks
When a current of 4.0A passes through a certain resistor for 10 minutes, 2.88 × 10J of heat are produced. Calculate:
  1. The power of the resistor.
  2. The voltage across the resistor.
Answer
Given: I = 4 amp, t = 10min = 10 × 60 = 600 sec, H = 2.88 × 104J

  1. We have

K = I2RT

28800 = 42 × R × 600

R = 3 ohm

We know that

P = I2 × R

= 42 × 3

P = 48W

  1. V = ?

We know that

V = IR

V = 4 × 3

V = 12V

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Question 53 Marks
What would be the effect on the resistance of a metal wire of:
  1. Increasing its length?
  2. Increasing its diameter?
  3. Increasing its temperature?
Answer
  1. Resistance will increase.
  2. Resistance will decrease.
  3. Resistance will increase.
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Question 63 Marks
What will be the resistance of a metal wire of length 2 metres and area of cross-section 1.55 × 10m , if the resistivity of the metal be 2.8 × 10 m?
Answer
I = 2m
A = 1.55 × 10-6 m2
P = 2.8 × 10-8m
$\text{R}=\text{P}\frac{\text{I}}{\text{A}}$
$=2.8\times10^{-8}\times\frac{2}{1.55\times10^{-6}}$
$0.036\Omega$
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Question 73 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 83 Marks
What is the resistance between A and B in the figure given below?
Answer
The three resistance of 20 ohm, 10 ohm and 20 ohm on the extreme right side are in series.
So, the resultant of these three resistance = 20 + 20 + 10 = 50 ohms.
This 50 ohms is in parallel with 30 ohms. so resultant of these two will be
$\frac{1}{\text{R}}=\frac{1}{30}+\frac{1}{50}$
$\frac{1}{\text{R}}=\frac{80}{1500}$
R = 18.75 ohms
Now, the resistance 10 ohms, 18.75 ohms and 10 ohms are in series.
Therefore, resultant resistance = 18.75 + 10 + 10 = 38.75 ohms.
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Question 93 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 103 Marks
What is, highest, and lowest, resistance which can be obtained by combining Com resistors having the following resistances?
$4 Ω, 8 Ω, 12 Ω, 24 Ω$
Answer
For obtaining the highest resistance by combining the given resistance, we must connect them in series.
We get,
R = 4 + 8 + 12 + 24 = 48 ohms
For obtaining the lowest resistance by combining the given resistance, we must connect them parallel.
We get,
$\frac{1}{\text{R}}=\frac{1}{4}+\frac{1}{8}+\frac{1}{12}+\frac{1}{24 }$
On solving we get, R = 2 ohms.
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Question 113 Marks
What is a voltmeter? How is a voltmeter connected in the circuit to measure the potential difference between two points. Explain with the help of a diagram.
Answer
A voltmeter is a device which is used to measure the potential difference between two points in an electric circuit. Voltmeter is always connected in parallel across the two points where the potential difference is to be measured.
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Question 123 Marks
What is an ammeter? How is it connected in a circuit? Draw a diagram to illustrate your answer.
Answer
Ammeter is a device used for the measurement of electric current. It is always connected in series with the circuit in which the current is to be measured.
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Question 133 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 143 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 153 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 163 Marks
Two exactly similar electric lamps are arranged (i) in parallel, and (ii) in series. If the parallel and series combination of lamps are connected to 220V supply line one by one, what will be the ratio of electric power consumed by them?
Answer
Let resistance of each lamp = R ohms.
Case1: Parellel connection
Resultant resistance $=\frac{1}{\frac{1}{\text{R}}+\frac{1}{\text{R}}}=\frac{\text{R}}{2}$
Electric power consumed $\text{P}_1=\frac{\text{V}_2}{\text{R}}=\frac{220^2}{\frac{\text{R}}{2}}=\frac{96800}{\text{R}}$
Case2: Series connection
Resultant resistance = R + R = 2R
Electric Power consumed $\text{P}_2=\frac{\text{V}^2}{2\text{R}}=\frac{24200}{\text{R}}$
$\therefore\frac{\text{P}_1}{\text{P}_2}=\frac{\frac{96800}{\text{R}}}{\frac{24200}{\text{R}}}=\frac{4}{1}$ 
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Question 173 Marks
Three 2V cells are connected in series and used as a battery in a circuit.
  1. What is the p.d. at the terminals of the battery?
  2. How many joules of electrical energy does 1C gain on passing through
  1. One cell.
  2. All three cells?
Answer
  1. 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.
  1.  
  1. Given: p.d. = 2V, Charge moved = 1C

We know that

Work done = p.d × charge moved

= 2 × 1

Work done = 2 joules

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

Work done = p.d × charge moved

= 6 × 1

Work done = 6 joule.

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Question 183 Marks
The resistors R1 , R2 , R3 and R4 in the figure given below are all equal in value.

What would you expect the voltmeters A, B and C to read assuming that the connecting wires in the circuit have negligible resistance?
Answer
Req = R + R + R + R = 4R ohm
Total current in the circuit, $\text{I}=\frac{\text{V}}{\text{R}}=\frac{12}{4\text{R}}=\frac{3}{\text{R}}$
Reading of voltmeter A = Voltage across $\text{R}_1=\text{I}\times\text{R}_1=\frac{3}{\text{R}}\times\text{R}=3\text{V}$
Reading of voltmeter B = Voltage across $\text{R}_2=\text{I}\times\text{R}_2=\frac{3}{\text{R}}\times\text{R}=3\text{V}$
Reading of voltmeter C = Voltage across the series combination of R3 and R4 $=\text{I}\times(\text{R}_3+\text{R}_4)=\frac{3}{\text{R}}\times2\text{R}=6\text{V}$
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Question 193 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:

  1. Electric wires.
  2. Handle for soldering iron, and
  3. Solar cells?
Give reasons for your choices.

Answer
  1. Material Q with resistivity 2.63 × 10-8 ohm-m can be used for making electric wires because it has very low resistivity.
  2. Material R with resistivity 1.0 × 1015 ohm-m can be used for making handle of soldering iron because it has very high resistivity.
  3. 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 203 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:
  1. Heating element of electric iron.
  2. Connecting wires of electric iron.
  3. 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 213 Marks
The electrical resistivities of four materials A, B, C and D are given below:
$\text{A}\ -110\times10^{-8}\Omega\text{ m}$
$\text{B}-\ 1.0\times10^{10}\Omega\text{ m}$
$\text{C}-\ 10.0\times10^{-8}\Omega\text{ m}$
$\text{D}-\ 2.3\times10^{3}\Omega\text{ m}$
Which material is:
  1. Good conductor.
  2. Resistor.
  3. Insulator, and
  4. Semiconductor
Answer
  1. Good conductor = C (10 × 10-8 ohm).
  2. Resistor = A (110 × 10-8 ohm).
  3. Insulator = B (1 × 1010 ohm).
  4. Semiconductor = D (2.3 × 103 ohm).
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Question 223 Marks
The electrical resistivities of five substances A, B, C, D and E are given below:
$\begin{matrix}\text{B}&110\times10^{-8}\Omega\text{ m}\\\text{C}&2.60\times10^{-8}\Omega\text{ m}\\\text{D}&10.0\times10^{-8}\Omega\text{ m}\\\text{E}&1.70\times10^{-8}\Omega\text{ m}\end{matrix}$
Answer
  1. E is best conductor of electricity due to its least electrical resistivity.
  2. C, because its resistivity is lesser than that of A.
  3. B, because it has the highest electrical resistivity.
  4. C and E, because of their low electrical resistivities.
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Question 233 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.
  1. Which circuit would have the highest voltage across each bulb?
  2. In which circuit would the bulbs be brighter?
  3. In which circuit, if one bulb blows out, all others will stop glowing
  4. Which circuit would have less current in it?
Answer
  1. Parallel Circuit

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

  1. Parallel Circuit

because all have higher voltage across them.

  1. Series Circuit

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

  1. Series Circuit

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

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Question 243 Marks
State three factors on which the heat produced by an electric current depends. How does it depend on these factors?
Answer
Heat produced by an electric current depends on the following factors:
  1. Heat produced is directly proportional to square of current.
  2. Heat produced is directly proportional to resistance.
  3. Heat produced is directly proportional to the time for which current flows.
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Question 253 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:
  1. Square of current.
  2. Resistance of wire.
  3. Time for which current is passed.
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Question 263 Marks
Show how you would connect two 4 ohm resistors to produce a combined resistance of:
  1. 2 ohms.
  2. 8 ohms.
Answer
By connecting in parallel: Since equivalent resistance will be
$\frac{1}{\text{R}}=\frac{1}{4}+\frac{1}{4}=\frac{2}{4}=\frac{1}{2}$
Therefore. R = 2 ohm
By connecting in series: since equilaent resistance will be R = 4 ohm + 4 ohm = 8 ohm.
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Question 273 Marks
In which of the following cases more electrical energy is consumed per hour?
  1. A current of 1 ampere passed through a resistance of 300 ohms.
  2. A current of 2 amperes passed through a resistance of 100 ohms.
Answer
  1. R = 300 ohm, I = 1A, t = 1h

P = I2 R = 12 × 300 = 300W

E = P × t = 300 × 1 = 300Wh

  1. 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 283 Marks
In the circuit shown below, the voltmeter reads 10V.

  1. What is the combined resistance?
  2. What current flows?
  3. What is the p.d. across $ 2 Ω$ resistor?
  4. What is the p.d. across $3 Ω$ resistor?
Answer
I = 6V
$\text{R}_1=2\Omega,\text{R}_2=3\Omega$
Combined resistance, Rtot = 2 + 3 = 5
$\text{I}=\frac{\text{V}}{\text{R}_\text{tot}}=\frac{10}{5}=2\text{A}$
p.d. across $2\Omega$ resistor = I × R1 = 2 × 2 = 4V.
p.d. across $3\Omega$ resistor = I × R2 = 2 × 3 = 6V.
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Question 293 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 303 Marks
In 10s, a charge of 25C leaves a battery, and 200j of energy are delivered to an outside circuit as a result.
  1. What is the p.d. across the battery?
  2. What current flows from the battery?
Answer
t = 10s
Q = 25C,
Energy deliverd = work done = 200J
$\text{p}.\text{d}.=\frac{\text{Work done}}{\text{Charge moved}}=\frac{200}{25}=8\text{V}$
$\text{I}=\frac{\text{Q}}{\text{t}}=\frac{25}{10}=2.5\text{A}$
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Question 313 Marks
If the charge on an electron is 1.6 × 10 coulombs, how many electrons should pass through a conductor in 1 second to constitute 1 ampere current?
Answer
We know that
$\text{I}=\frac{\text{Q}}{\text{T}}$
$\Rightarrow1\text{A}=\frac{\text{Q}}{1\text{s}}$
$\Rightarrow\text{Q}=1\text{C}$
Now, when charge is 1.6 × 10-19 Coulamb, number of electrone = 1
When charge is 1 Coulamb, number of electrone $=\frac{1}{1.6\times10^{-19}}=0.625\times10^{19}=625\times10^{18}$
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Question 323 Marks
If a potential difference of 10V causes a current of 2A to flow for 1 minute, how much energy is transferred?
Answer
Given: p.d. = 10V, I = 2amp, t = 1 min = 60s.
We know that:
$\text{I} = \frac{\text{Q}}{\text{T}}$
Thus, Q = I × t.
Q = × 60.
Q = 120C.
Work done = p.d. × charge moved
Work done = 120 × 10J
Work done = 1200J.
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Question 333 Marks
How will you connect three resistors of resistances $2Ω, 3Ω$ and $6Ω $ obtain a total n si stance of:
  1. $4 Ω,$
  2. $1Ω$
Answer
  1. Connects 2 ohms resistor in series with a parallel combinations of 3 ohms and 6 ohms.
  2. Connects 2 ohms, 3 ohms, and 6 ohms in parallel.
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Question 343 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 353 Marks
How does the resistance of a wire change when:
  1. Its length is tripled?
  2. Its diameter is tripled?
  3. Its material is changed to one whose resistivity is three times?
Answer
$\text{R}=\rho\frac{\text{I}}{\text{A}}$

  1. I → 3I

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

Resistance gets tripled.

  1. 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 $\frac{1}{9}\text{th}.$

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

$\rho\rightarrow3\rho$

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

Resistance becomes 3 times.

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Question 363 Marks
How does the resistance of a conductor depend on:
  1. Length of the conductor?
  2. Area of cross-section of the conductor?
  3. Temperature of the conductor?
Answer
  1. Resistance of a conductor increases (or decreases) with increase (or decrease) in the length of the conductor.
  2. Resistance of a conductor decreases (increases) with increase (decrease) in the area of cross-section of the conductor.
  3. Resistance of a conductor increases on raising the temperature and decreases on lowering the temperature.
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Question 373 Marks
Give three reasons why different electrical appliances in a domestic circuit are connected in parallel.
Answer
Different electrical appliances in a domestic circuit are connected in parallel because of the following advantages:
  1. If one electrical appliance stops working due to some defect, then all other appliances keep working properly.
  2. Each electrical appliance has its own switch due to which it can be turned on or turned off independently, without affecting other appliances.
  3. Each electrical appliance gets the same voltage as that of the power supply line.
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Question 383 Marks
Give three reasons why different electrical appliances in a domestic circuit are connected in parallel.
Answer
Different electrical appliances in a domestic circuit are connected in parallel because of the following advantages:
  1. If one electrical appliance stops working due to some defect, then all other appliances keep working properly.
  2. Each electrical appliance has its own switch due to which it can be turned on or turned off independently, without affecting other appliances.
  3. Each electrical appliance gets the same voltage as that of the power supply line.
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Question 393 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 403 Marks
Draw circuit symbols for:
  1. Fixed resistance.
  2. Variable resistance.
  3. A cell.
  4. A battery of three cells.
  5. An open switch.
  6. A closed switch.
Answer
  1. Fixed resistance


  1. Variable Resistance.


  1. Cell.


  1. Battery of three cells.


  1. Open Switch.


  1. Closed Switch.

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Question 413 Marks
Distinguish between resistance and resistivity.
Answer
  1. Resistance is the property of the conductor, while resistivity is the property of the material of the conductor.
  2. 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.
  3. Resistivity of a substances depends on the nature of the substance and temperature.
  4. 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 423 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 433 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 443 Marks
Consider the circuit given below where A, B and C are three identical light bulbs of constant resistance.

  1. List the bulbs in order of increasing brightness.
  2. If C burns out, what will be the brightness of A now compared with before?
  3. If B burns out instead, what will be the brightness of A and C compared with before?
Answer
  1. 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.
  2. A gets the same voltage as before, so its brightness remains the same.
  3. 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 453 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 463 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 473 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 483 Marks
Calculate the power used in the $2Ω$ resistor in each of the following circuits:
  1. A 6V battery in series with $1Ω$ and $2Ω$ resistors.
  2. A 4V battery in parallel with $12Ω$ and $2Ω$ resistors.
Answer
  1. 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

  1. 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 493 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 503 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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