Chapter 12 – Electricity Solutions
Question - 11 : - Use the data in the table given below and answer the following questions.
Answer - 11 : -
| Material | Resistivity |
| Conductors | Silver | 1.60 × 10–8 |
| Copper | 1.62 × 10–8 |
| Aluminium | 2.63 × 10–8 |
| Tungsten | 5.20 × 10–8 |
| Nickel | 6.84 × 10–8 |
| Iron | 10.0 × 10–8 |
| Chromium | 12.9 × 10–8 |
| Mercury | 94.0 × 10–8 |
| Manganese | 1.84 × 10–6 |
| Alloys | Constantan | 49 × 10–6 |
| Manganin | 44 × 10–6 |
| Nichrome | 100 × 10–6 |
| Insulators | Glass | 1010 – 1014 |
| Hard rubber | 1013 – 1016 |
| Ebonite | 1015 – 1017 |
| Diamond | 1012 – 1013 |
| Paper (dry) | 1012 |
a. Which among iron and mercury is a better conductor?
b. Which material is the best conductor?
Answer
a. Iron is a better conductor than mercury because the resistivity of mercury is more than the resistivity of iron.
b. Among all the materials listed in the table, silver is the best conductor because the resistivity of silver is lowest among all, i.e., 1.60 × 10–8.
Question - 12 : - Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series.
Answer - 12 : - A battery of three cells of 2 V each equals to battery of potential 6 V. The circuit diagram below shows three resistors of resistance 12 Ω, 8 Ω and 5 Ω connected in series along with a battery of potential 6 V.
Question - 13 : - Redraw the circuit of Question 1, putting in an ammeter to measure the current through-
Answer - 13 : - the resistors and a voltmeter to measure the potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter?
Answer
An ammeter should always be connected in series with resistors while the voltmeter should be connected in parallel to the resistor to measure the potential difference as shown in the figure below.
Using Ohm’s Law, we can obtain the reading of the ammeter and the voltmeter.
The total resistance of the circuit is 5 Ω + 8 Ω +12 Ω = 25 Ω.
We know that the potential difference of the circuit is 6 V, hence the current flowing through the circuit or the resistors can be calculated as follows:
I = V/R = 6/25 = 0.24A
Let the potential difference across the 12 Ω resistor be V1.
From the obtained current V1 can be calculated as follows:
V1 = 0.24A × 12 Ω = 2.88 V
Therefore, the ammeter reading will be 0.24 A and the voltmeter reading be 2.88 V.
Question - 14 : - Judge the equivalent resistance when the following are connected in parallel – (a) 1 Ω and 106 Ω, (b) 1 Ω, 103 Ω, and 106 Ω.
Answer - 14 : - (a) When 1 Ω and 106 are connected in parallel, the equivalent resistance is given by
Therefore, the equivalent resistance is 1 Ω.
(b) When 1 Ω, 103 Ω, and 106 Ω are connected in parallel, the equivalent resistance is given by
Therefore, the equivalent resistance is 0.999 Ω.
Question - 15 : - An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source.
Answer - 15 : - What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?
Answer
The electric lamp, the toaster and the water filter connected in parallel to a 220 V source can be shown as using a circuit diagram as follows:
The equivalent resistance of the resistors can be calculated as follows:
The resistance of the electric iron box is 31.25 Ω.
Question - 16 : - What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
Answer - 16 : - When the electrical devices are connected in parallel there is no division of voltage among the appliances. The potential difference across the devices is equal to supply voltage. Parallel connection of devices also reduces the effective resistance of the circuit.
Question - 17 : - How can three resistors of resistances 2 Ω, 3 Ω, and 6 Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω?
Answer - 17 : - (a) The circuit diagram below shows the connection of three resistors
From the circuit above, it is understood that 3 Ω and 6 Ω are connected in parallel. Hence, their equivalent resistance is given by
The equivalent resistor 2 Ω is in series with the 2 Ω resistor. Now the equivalent resistance can be calculated as follows:
Req= 2 Ω +2 Ω = 4 Ω
Hence, the total resistance of the circuit is 4 Ω.
(b) The circuit diagram below, shows the connection of three resistors.
From the circuit, it is understood that all the resistors are connected in parallel. Therefore, their equivalent resistance can be calculated as follows:
The total resistance of the circuit is 1 Ω.
Question - 18 : - What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?
Answer - 18 : -
(a) If the four resistors are connected in series, their total resistance will be the sum of their individual resistances and it will be the highest. The total equivalent resistance of the resistors connected in series will be 4 Ω + 8 Ω + 12 Ω + 24 Ω = 48 Ω.
(b) If the resistors are connected in parallel, then their equivalent resistances will be the lowest.
Their equivalent resistance connected in parallel is
Hence, the lowest total resistance is 2 Ω.
Question - 19 : - Why does the cord of an electric heater not glow while the heating element does?
Answer - 19 : - The heating element of an electric heater is made of an alloy which has a high resistance. When the current flows through the heating element, the heating element becomes too hot and glows red. The cord is usually made of copper or aluminum which has low resistance. Hence the cord doesn’t glow.
Question - 20 : - Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
Answer - 20 : -
The heat generated can be computed by Joule’s law as follows:
H = VIt
where,
V is the voltage, V = 50 V
I is the current
t is the time in seconds, 1 hour = 3600 seconds
The amount of current can be calculated as follows: