1 ohm is the resistance offered by a conductor of uniform cross-sectional area when 1 A current flows through it at a voltage of 1 volt . 1 ohm is defined using Ohm's law. According to Ohm's law, V = I R ⇒ R = V I .

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