Class 10 CBSE last 10 year questions With Solution from electricity from CBSE question paper (2020-2010)

Q1 At the time of short circuit, the electric current in the circuit:

(a) vary continuously (b) does not change (c) reduces substantially (d) increases heavily [1]

Answer: (D) increases heavily

Q2 Two bulbs of 100 W and 40 W are connected in series. The current through the 100 W bulb is 1A. The current through the 40W bulb will be:

(a) 0.4A (b) 0.6A (c) 0.8A (d) 1A [1]

Answer: (D) 1A (Because it is connected in series and we know that in series current remains same)

Q3 A V-I graph for a nichrome wire is given below. What do you infer from this graph? Draw a labelled circuit diagram to obtain such a graph. [3]

Answer: The V-I graph of the nichrome wire shows a straight line. It means that the resistance of the wire remains constant when the current supply is changed. It follows the ohm's low and therefore, it is considered as ohmic conductor.
It follows the ohm's law $\to V = I R$
$R = \frac{V}{I}$

Q4 (a) Write the mathematical expression for Joule’s law of heating:

(b) Compute the heat generated while transferring 96000 coulomb of charge in two hours through a potential difference 40 V. [3]

Answer: (a) H square is equal to I square into R into T. This is called joule's law of heating.The devices which works on this principle are Electric Heater, Immersion water Heater, Electric bulb,Fuse wire etc.

(b) Given :

Amount of charge transferred $= 96000 C$

Time taken $= 2 h r s = 2 \times 60 \times 60 s e c$
$= 7200 s e c$

Potential difference $= 40 V$

Heat generated $= V \times i \times t$

and we know that ; $i = \frac{Q}{t}$

So, $H = V Q$
$= 40 \times 96000$
$= 3.84 \times 1 0^{6} J$
2019

Q5 Write the function of voltmeter in an electric circuit.

ANSWER: The function of the voltmeter in a circuit is to measure the voltage drop across any appliance. The voltmeter is always connected parallel in the circuit.

Q6 (a) With the help of a suitable circuit diagram prove that the reciprocal of the equivalent resistance of a group of resistances joined in parallel is equal to the sum of the reciprocals of the individual resistances.

(b) In an electric circuit, two resistors of 12 ohm each are joined in parallel to a 6 V battery. Find the current drawn from the battery.

Answer: (a) It is observed that total current $I$ is equaql to the sum not seperate current .

$I=I_{1}+I_{2}+I_{3}$ ___(i)

Let $R_{P}$ be the equivalent resistance of he parallel combination of resistance.

By applying Ohm's law , $I=\frac{V}{R_{P}}$ , _____(ii)

From (1) and (ii)

$\frac{V}{R_{P}}=\frac{V}{R_{1}}+\frac{V}{R_{2}}+\frac{V}{R_{3}}$

Cancel V from both sides

$\frac{1}{R_{P}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}$

Hence,if u resistance are connected in parallel, then the equivalent resistance of the circuit -

$\frac{1}{R_{eq}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+----\frac{1}{R_{n}}$

(b) Given, Two resistors of $12\Omega$ connected in parallel.

$V=6V$

$\therefore \frac{1}{R_{eq}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}$

$\frac{1}{R_{eq}}=\frac{1}{12}+\frac{1}{12}=\frac{2}{12}$

According to Ohm's law,

$V=IR$

$6=I\times 6$

$\frac{6}{6}=I$

$I=1\; amphere$

Q7 An electric lamp of resistance 20 0 and a conductor of resistance 4 0 are connected to a 6 V battery as shown in the circuit. Calculate:

(a) The total resistance of the circuit
(b) The current through the circuit
(c)The potential difference across the (i) Electric lamp and
(ii) Conductor, and
(d) Power of the lamp.

Answer

a). the total resistance of the circuit = $20 Ω + 4 Ω = 24 Ω$

b). The current through the circuit= current through the bulb= current through the conductor.
The current through the circuit= $\frac{v o l t a g e a p p l i e d}{t h e v o l t a g e o f t h e b a t t e r y} .$
the current through the circuit $= \frac{6}{2 4} a m p$

The current through the circuit = $0.25 a m p$

c). The potential difference across the lamp and the conductor.
(i) The potential difference across the electrical lamp $= 0.25 \times 20 = 5 v o l t$
(ii) The potential difference conductor $= 0.25 \times 4 = 1 v o l t$

d). Power can be calculated as $\frac{v o l t a g e a c r o s s t h e l a m p}{c u r r e n t t h r o u g h t h e l a m p}$
power= $\frac{5}{0 . 2 5} \to 20 W$ .

Q8 While studying the dependence of potential difference ( V) across a resistor on the current (I) passing through it, in order to determine the resistance of the resistor, a student took 5 readings for different values of current and plotted a graph between V and t. He got a straight line graph passing through the origin. What does the straight line signify? Write the method of determining the resistance of the resister using this graph.

Answer

When the student took 5 readings for different values of R and the student gets a straight line graph.
This verifies Ohm's law.
Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points
So, if you plot a graph of current against voltage you will get:

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

Q9 What would you suggest to a student if while performing an experiment he finds that the pointer/needle of the ammeter and voltmeter do not coincide with the zero marks on the scales when the circuit is open? No extra ammeter/voltmeter is available in the laboratory.
Answer: This is called the zero error of the scale of ammeter or voltmeter. If there is a zero error, then this error is subtracted from the value that depicts when the circuit is closed, otherwise, the accurate current or potential difference will not be recorded.

2018

Q10 Show how would you join three resistors, each of resistance 9 ohm so that the equivalent resistance of the combination is (i) 13 ohm (ii) 6 ohm ?
Ans. (i) two 9 ohm resistors are connected in parallel and one in series
(ii) 2 resistors connected in series =9+9= 18 ohm

Q11 (a) Write Joule’s law of heating. (b) Two lamps, one rated 100 W; 220 V, and the other 60 W; 220 V, are connected in parallel to electric mains supply. Find the current drawn by two bulbs from the line, if the supply voltage is 220 V.
Answer
a) Joule's law of heating can be defined as: when current passes through a conductor, heat is generated. It is also known as Ohmic heating or resistive heating.
P = I² R = V²/R
Where, P is in watts, I is in Amperes, R is in ohms and V is volts.
b) firstly we would find the resistance of the lamps.
P₁ = 100 W
V₁ = 220 V
R ₁ = V₁²/P₁ = 484 Ω
P₂ = 60 W
V₂ = 220 V
R₂ = V₂²/P₂ = 806.67 Ω
The lamps are connected in parallel. Equivalent resistance =
R = (R₁×R₂)/(R₁+R₂) = (484 Ω×806.67 Ω)( 484 Ω+806.67 Ω) = 302.5 Ω
From Ohm's law, current drawn is given by:
I = V /R = 220 V/ 302.5Ω = 0.727 A

Q12 (a) List the factors on which the resistance of a conductor in the shape of a wire depends.
(b) Why are metals good conductors of electricity whereas glass is a bad conductor of electricity? Give reason.
(c) Why are alloys commonly used in electrical heating devices ? Give reason.

Answer:
(a) Resistance of a conductor depends directly on its length and is inversely proportional to the area of cross-section.
(b) Metals have free electrons and they can move and conduct electricity, whereas glass does not allow electrons and charges to flow freely as it is an insulator.
(c) The resistivity of an alloy is generally higher than that of its constituent metals. Alloys do not oxidise (burn) readily at higher temperatures. Therefore, conductors of electric heating devices, such as toasters and electric irons, are made of an alloy rather than pure metal.

2017-16-15-14-13-12-11-  No Questions
2010
Q13 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 the relation known? How much heat will an instrument of 12 W produce in one minute if it is connected to a battery of 12 V?(5)
Solution: Let us take a resistor of resistance R. Let the current flowing through this resistor is equal to I and the potential difference across it is equal to V. Suppose in time t, Q amount of charge flows through the resistor Work done in moving this charge, W = VQ … (1) According to the definition of electric current, Q I t  Q I t  Putting this in equation (1), W = V × I × t This work done is dissipated as heat. Hence, heat produced, H = W = VIt H = VIt … (2) According to Ohm’s law, V = IR. Putting this in equation (2), H = IR × It H = I2Rt This relation is known as Joule’s law of heating Numerical: Power, P = 12 W Potential difference, V = 12 volt Time duration, t = 1 min = 60 s H P t  H P t  = 12 W × 60 s = 720 J The heat generated by the instrument is 720 J.

Q14 Explain with the help of a labeled circuit diagram how you will find the resistance of a combination of three resistors, of resistance R1, R2 and R3 joined in parallel. Also mention how you will connect the ammeter and the voltmeter in the circuit when measuring the current in the circuit and the potential difference across one of the three resistors of the combination. (5)

Answer

Suppose total current flowing in the circuit is I. then the current passing through the resistance $R_{1}$ will be $I_{1} .$ current passing through resistance $R_{2}$ will be $I_{2}$ and current passing through resistance $R_{3}$ will be $I_{3} .$
Total current $= I = I_{1} + I_{2} + I_{3}$
Let resultant resistance of this parallel combination is R. By applying the ohm's law to the whole circuit , we get that
$I_{1} = V / R_{1}$
$I_{2} = V / R_{2}$
$I_{3} = V / R_{3}$
Putting these eqs. in the above once , we get
$V / R = V / R_{1} + V / R_{2} + V / R_{3}$
$1 / R = 1 / R_{1} + 1 / R_{2} + 1 / R_{3}$
If two resistance are connected in parallel, then the resultant resistance will be
$1 / R = 1 / R_{1} + 1 / R_{2} + 1 / R_{3}$

(b) If switch is open then only upper two resistances (connected in parallel) are in the circuit.
Effective resistance is $1 / R_{e q} = 1 / R + 1 / R = 2 / R$
$R_{e q} = R / 2$
So the current $= I = V / (R / 2) = 0.6 A$ (given)
$V / R = 0.3 A$
When the switch closes , the third resistance also comes in the circuit. The effective resistance of the circuit becomes $R / 3$
Hence , current $= I = V / (R / 3) = 3 (V / R) = 3 \times 0.3 = 0.9 A$

GRADEUP 1

Class 10 CBSE last 10 year questions With Solution from electricity from CBSE question paper (2020-2010)

Answer: (D) 1A (Because it is connected in series and we know that in series current remains same)

Answer: The V-I graph of the nichrome wire shows a straight line. It means that the resistance of the wire remains constant when the current supply is changed. It follows the ohm's low and therefore, it is considered as ohmic conductor.
It follows the ohm's law $\to V = I R$
$R = \frac{V}{I}$

Amount of charge transferred $= 96000 C$

Time taken $= 2 h r s = 2 \times 60 \times 60 s e c$
$= 7200 s e c$

Potential difference $= 40 V$

Heat generated $= V \times i \times t$

and we know that ; $i = \frac{Q}{t}$

So, $H = V Q$
$= 40 \times 96000$
$= 3.84 \times 1 0^{6} J$
2019

(a) The total resistance of the circuit
(b) The current through the circuit
(c)The potential difference across the (i) Electric lamp and
(ii) Conductor, and
(d) Power of the lamp.

Answer

Answer

Answer

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CBSE Last 10(ten) Year Electricity questions with solution Class 10th

Class 10 CBSE last 10 year questions With Solution from electricity from CBSE question paper (2020-2010)

Answer: (D) 1A (Because it is connected in series and we know that in series current remains same)

Answer: The V-I graph of the nichrome wire shows a straight line. It means that the resistance of the wire remains constant when the current supply is changed. It follows the ohm's low and therefore, it is considered as ohmic conductor.It follows the ohm's law →V=IRR=IV​

Amount of charge transferred =96000CTime taken =2hrs=2×60×60sec =7200secPotential difference =40VHeat generated =V×i×tand we know that ; i=tQ​So, H=VQ =40×96000 =3.84×106J2019

(a) The total resistance of the circuit(b) The current through the circuit(c)The potential difference across the (i) Electric lamp and (ii) Conductor, and (d) Power of the lamp.

Answer

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Answer: The V-I graph of the nichrome wire shows a straight line. It means that the resistance of the wire remains constant when the current supply is changed. It follows the ohm's low and therefore, it is considered as ohmic conductor.
It follows the ohm's law $\to V = I R$
$R = \frac{V}{I}$

Amount of charge transferred $= 96000 C$

Time taken $= 2 h r s = 2 \times 60 \times 60 s e c$
$= 7200 s e c$

Potential difference $= 40 V$

Heat generated $= V \times i \times t$

and we know that ; $i = \frac{Q}{t}$

So, $H = V Q$
$= 40 \times 96000$
$= 3.84 \times 1 0^{6} J$
2019

(a) The total resistance of the circuit
(b) The current through the circuit
(c)The potential difference across the (i) Electric lamp and
(ii) Conductor, and
(d) Power of the lamp.