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

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The resistance of a resistor is reduced to half of its initial value. If other parameters of the electrical circuit remain unaltered, the amount of heat produced in the resistor will become:

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Important Questions on Electricity

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EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR SCIENCE>Chapter 12 - Electricity>CBSE-2023 (Delhi - I)>Q 38

Why is an alternating current (A.C.) considered to be advantageous over direct current (D.C.) for the long distance transmission of electric power?

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EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR SCIENCE>Chapter 12 - Electricity>CBSE-2023 (Delhi - I)>Q 39

How is electric current related to the potential difference across the terminals of a conductor? Draw a labelled circuit diagram to verify this relationship.

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EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR SCIENCE>Chapter 12 - Electricity>CBSE-2023 (Delhi - I)>Q 40

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.

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10th CBSE

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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:

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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.

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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.

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10th CBSE

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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.

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10th CBSE

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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.

The resistance of a resistor is reduced to half of its initial value. If other parameters of the electrical circuit remain unaltered, the amount of heat produced in the resistor will become:

Important Questions on Electricity

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