Calculate the equivalent resistance of the following network:

Let R p be the equivalent resistance of resistors connected in parallel. Then, 1 R p = 1 20 + 1 20 ⇒ 1 R p = 1 + 1 20 = 2 20 = 1 10 R p = 10 Ω The equivalent circuit becomes Here, the two, 10 Ω resistors are connected in series. ∴ The equivalent resistance of the circuit, R = 10 Ω + 10 Ω = 20 Ω Hence, the equivalent resistance of the given circuit is 20 Ω .

Three resistors R 1 , R 2 and R 3 are connected in parallel and the combination is connected to a battery, ammeter, voltmeter and key. Draw suitable circuit diagram and obtain an expression for the equivalent resistance of the combination of the resistors.

Let R 1 , R 2 & R 3 be the three resistances connected in parallel to one another and let R be the equivalent resistance of the circuit. A battery of V volts has been applied across the ends of this resistance combination. When the switch of the key is closed, a current I flows in the circuit. According to Ohm's law, I = V R I 1 = V R 1 I 2 = V R 2 I 3 = V R 3 I = I 1 + I 2 + I 3 - - - - ( i ) Substituting the values of I 1 , I 2 & I 3 in the equation ( i ) , we get V R = V R 1 + V R 2 + V R 3 V ( 1 R ) = V ( 1 R 1 + 1 R 2 + 1 R 3 ) 1 R = 1 R 1 + 1 R 2 + 1 R 3

Consider the given circuit and find the current flowing in the circuit and potential difference across the 15 Ω resistor when the circuit is closed.

Given, R 1 = 5 Ω , R 2 = 10 Ω , R 3 = 15 Ω Total resistance, R = R 1 + R 2 + R 3 [The three resistors in the circuit are connected in series] ⇒ R = 5 Ω + 10 Ω + 15 Ω = 30 Ω Potential difference, V = 30 V We have to find the current ( I ) flowing in the circuit. From Ohm’s law, I = V R ⇒ I = 30 V 30 Ω = 1 A Thus, the current flowing in the circuit = 1 A The potential difference across 15 Ω resistor, V 3 = I R 3 = 1 A × 15 Ω = 15 V Thus, the potential difference across 15 Ω resistor is 15 V .

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 .

From Joule's Law of heating, I = P V . First bulb, P 1 = 100 W , V = 220 V I 1 = P 1 V = 100 220 = 0 . 45 A Second bulb, P 2 = 60 W , V = 220 V I 2 = P 2 V = 60 220 = 0 . 27 A So, total current is I 1 + I 2 = 0 . 45 A + 0 . 27 A = 0 . 72 A .

Write Joule's law of heating.

According to Joule's law of heating, the heat generated in the resistance wire is directly proportional to the square of the current flowing through the wire H ∝ I 2 . directly proportional to the resistance of the wire at constant current H ∝ R . directly proportional to the time of flow of current H ∝ t . So, H = I 2 R t .

How will you infer with the help of an experiment that the same current flows through every part of a circuit containing three resistors in series connected to a battery?

When two or more resistors are joined end to end, the same current flows through each one of them when a potential difference is applied across the combination, then the resistors are said to be connected in series circuit. Let three resistors R 1 , R 2 & R 3 are connected in series. They are also connected with a battery, an ammeter and a key as shown in the figure. When we close the key, the current starts flowing through the circuit. Then we have to take the reading of ammeter. Now we have to change the position of the ammeter to anywhere in between the resistors and take the reading of it. We will observe that in both cases reading of the ammeter will be the same. This shows that the same current flows through every part of the above circuit.

Show how would you join three resistors, each of resistance 9 Ω , so that the equivalent resistance of the combination is 6 Ω .

Resistance is a measure of the opposition to current flow in an electrical circuit. To get an equivalent resistance of 6 Ω , the resistances should be connected as shown in the figure given below: Resistance of the series resistors is R = R 1 + R 2 R = 9 + 9 = 18 Ω Now, the total resistance is the sum of this effective resistance and the resistance of the resistor in parallel. That is, 1 R = 1 18 Ω + 1 9 Ω = 3 18 = 1 6 Ω Therefore , R = 6 Ω

Show how would you join three resistors, each of resistance 9 Ω , so that the equivalent resistance of the combination is 13 . 5 Ω .

Resistance is a measure of the opposition to current flow in an electrical circuit. To get an equivalent resistance of 13 . 5 Ω , the resistances should be connected as shown in the figure given below: Resistance of the parallel resistors is 1 R = 1 9 + 1 9 1 R = 2 9 T h e r e f o r e , R = 9 2 = 4 . 5 Ω Now, the total resistance is the sum of this effective resistance and the resistance of the resistor in series. That is, R = 9 Ω + 4 . 5 Ω = 13 . 5 Ω

MEDIUM

10th CBSE

IMPORTANT

Earn 100

Two $V - I$ graphs A and B for series and parallel combinations of two resistors are as shown. Giving reason state which graph shows (a) series, (b) parallel combination of the resistors.

Important Questions on Electricity

MEDIUM

10th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR SCIENCE>Chapter 12 - Electricity>CBSE-2023 (Delhi - II)>Q 41

Two LED bulbs of

10 W

and

5 W

are connected in series. If the current flowing through

5 W

bulb is

0.005 A

, the current flowing through

10 W

bulb is:

MEDIUM

10th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR SCIENCE>Chapter 12 - Electricity>CBSE-2023 (Outside Delhi - I)>Q 42

Three incandescent bulbs of $100 W$ each are connected in series in an electric circuit. In another set of three bulbs of the same wattage are connected in parallel to the same source. Now let one bulb in both the circuits get fused. Will the rest of the bulbs continue to glow in each circuit? Give reason.

HARD

10th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR SCIENCE>Chapter 12 - Electricity>CBSE-2023 (Outside Delhi - I)>Q 43

Three incandescent bulbs of $100 W$ each are connected in series in an electric circuit. In another circuit another set of three bulbs of the same wattage are connected in parallel to the same source. Will the bulb in the two circuits glow with the same brightness? Justify your answer.

HARD

10th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR SCIENCE>Chapter 12 - Electricity>CBSE-2023 (Outside Delhi - I)>Q 44

Let the resistance of an electrical device remain constant, while the potential difference across its two ends decreases to one fourth of its initial value. What change will occur in the current through it? State the law which helps us in solving the above stated question.

MEDIUM

10th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR SCIENCE>Chapter 12 - Electricity>CBSE-2023 (Outside Delhi - I)>Q 45

What is meant by resistance of a conductor? Define its SI unit.

HARD

10th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR SCIENCE>Chapter 12 - Electricity>CBSE-2023 (Outside Delhi - I)>Q 46

List two factors on which the resistance of a rectangular conductor depends.

HARD

10th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR SCIENCE>Chapter 12 - Electricity>CBSE-2023 (Outside Delhi - I)>Q 47

How will the resistance of a wire be affected if its
(1) length is doubled, and
(2) radius is also doubled?
Give justification for your answer.

MEDIUM

10th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR SCIENCE>Chapter 12 - Electricity>CBSE-2023 (Outside Delhi - II)>Q 48

Three resistors of

6 Ω, 4 Ω

and

4 Ω

are connected together so that the total resistance is

8 Ω

. Draw a diagram to show this arrangement and give reason to justify your answer.

Two V-I graphs A and B for series and parallel combinations of two resistors are as shown. Giving reason state which graph shows (a) series, (b) parallel combination of the resistors.

Important Questions on Electricity

Learn and Prepare for any exam you want !

Two $V - I$ graphs A and B for series and parallel combinations of two resistors are as shown. Giving reason state which graph shows (a) series, (b) parallel combination of the resistors.