|Material||Resistivity (Ω m)|
|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|
(alloy of Cu and Ni)
|49 � 10−6|
(alloy of Cu, Mn and Ni)
|44 � 10−6|
(alloy of Ni, Cr, Mn and Fe)
|100 � 10−6|
|Glass||1010 − 1014|
|Insulators||Hard rubber||1013 − 1016|
|Ebonite||1015 − 1017|
|Diamond||1012 − 1013|
(a) Resistivity of iron = 10.0 x 10-8 Ω
Resistivity of mercury = 94.0 x 10-8 Ω
Resistivity of mercury is more than that of iron. This implies that iron is a better conductor than mercury.
(b) It can be observed from Table 12.2 that the resistivity of silver is the lowest among the listed materials. Hence, it is the best conductor.
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.
Redraw the circuit of question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter?
The resistances are connected in series.
Ohm’s law can be used to obtain the readings of ammeter and voltmeter. According to Ohm’s law,
V = IR,
Potential difference, V = 6 V
Current flowing through the circuit/resistors = I
Resistance of the circuit, R = 5 + 8 + 12 = 25Ω
I = V/R = 6/25 = 0.24 A
Potential difference across 12 Ω resistor = V1
Current flowing through the 12 Ω resistor, I = 0.24 A
Therefore, using Ohm’s law, we obtain
V1 = IR = 0.24 x 12 = 2.88 V
Therefore, the reading of the ammeter will be 0.24 A.
The reading of the voltmeter will be 2.88 V.
Judge the equivalent resistance when the following are connected in parallel − (a) 1 Ω and 106Ω, (b) 1 Ω and 103Ω and 106Ω.
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 Vsource. 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?
According to Ohm's law,
V = IR
Current flowing through the circuit = I
7.04 A of current is drawn by all the three given appliances.
Therefore, current drawn by an electric iron connected to the same source of potential 220 V = 7.04 A
Let R' be the resistance of the electric iron. According to Ohm's law,