G1 G2
A B
(C) 1.0 (D) cannot be determined
Q.7 In the circuit shown in figure, the reading of ammeter A is
(A) 0.1 A (B) 0.2 A (C) 0.3 A (D) 0.4 A
Q.8 Two resistances X and Y are to be measured using an ammeter of resistance 0.5 and a voltmeter of resistance 20 K. It is known that X is in the range of a few ohms while Y is in the range of several thousand ohms.
Which of the following two connections would give minimum error in V
A
V
A
the measurements of resistances X and Y?
(A) circuit (a) for both X and Y (B) circuit (b) for both X and Y (C) circuit (a) for X and circuit (b) for Y (D) circuit (b) for X and circuit (a) for Y
ANSWERS
(1) (B) (2) (B) (3) (D) (4) (D)
(5) (C) (6) (C) (7) (D) (8) (C)
HEATING EFFECT OF CURRENT, JOULE'S LAW 1. Work Heat
or Work = a constant × heat or W = JH
Where J is known as Joule's constant or mechanical equivalent of heat. It is defined as that mechanical work which produces unit calorie of heat.
2. Expression for heat produced in a conductor due to current flow through it-(a) Let the potential difference between the points A and B of a conductor is V, on
account of which a current i flows through it.
A i B
R (b) As potential difference is the work done per unit test charge.
W = QV But Q = it
W = Vit.
But heat developed
W Vit i Rt2
H J J J Where R is the resistance of wire.
(c) If i is in ampere, V is in volt and R is in ohm, then W = Vit Joule = Vit × 107 ergs.
Heat developed H Vit
J calories. = Vit
4.2 = 0.24 Vit cals = 0.24 i2Rt cals.
Joule's laws of heating effects of current
(i) The heat developed in a conductor is given by H = i2Rt Joule = 0.24 i2Rt cals.
(ii) The amount of heat developed in a conductor, in a given time, is directly proportional to the square of the current.
i.e. H i2, when R and t are constants.
(iii)The amount of heat developed in a conductor by a given current in a given time is directly proportional to the resistance of the conductor. i.e. H R, when i and t are constants.
(iv) The amount of heat developed in a given conductor due to a given current is proportional to the time of flow of the current.
i.e. H t, when i and R are constants.
Electrical energy or work
If Q units of charge be carried between two points differing in potential by V, then electrical work done is W = Q × V Joule = Q × V × 107 ergs.
Power
The rate at which work is done is defined as power
Power = Q V t
= Vi = i2 R = V2
R Rated power (of an electric device)
The voltage of operation needed is known as the rated voltage and the corresponding power consumed is called as the rated power.
Suppose, R is the electrical resistance of an electric device, with Vr and Pr its is rated voltage and wattage respectively. Clearly,
2 r r
P V
R ... (1)
Now, if the device is subjected to a voltage of V, then, the power consumption P will be V2
P R ... (2)
The resistance R being the property of the electric device, is independent of the applied voltage.
Dividing eq. (2) by eq. (1)
2 r r
P V P
V
Thus, if the voltage be times the rated one, then the actual power consumption will be 2 times the rated one.
Thermal Equilibrium of a Heat Emitting Filament
Consider a filament of length L and cross-sectional area S (=r2 ) carrying a current i maintaining an equilibrium temperature T (in absolute scale). The thermal power generated in it is
P = i²R or V2/R ... (1) The heat energy dissipated (from Stefan Boltzman’s law) per unit time is
4 4
H (2 rL) [T T ]0 ... (2)
where = Stefen’s constant = 5.67 10 8W / m -K2 4, TT0 = temperature of the surrounding
and = emissivity of the surface of the filament.
Form thermal equilibrium, the rate of heat generated due to current = the rate of heat emission due to temperature difference between the filament and the surrounding.
2 4 4
0
V (2 rL) (T T )
R ... (3)
If 0 and be the resistivities of the material of the filament at 0°C and t°C respectively then
0(1 t)
where is the temperature coefficient of the material of the filament.
If T = (273 + t), then L 0(1 2t) L
R S r
(Ignoring any change in dimensions of the filament due to temperature rise]
Eq. (3) changes to
2 2
Commercially it is a device employed to save the different electrical appliances used in a house such as fan, bulb, T.V., tape recorder etc. in circumstances of abrupt increase of currents entering through the main supply.
The usual currents supposed to activate household appliances lies within 0-5.0 A.
An electric fuse consists of a simple wire made of an alloy of tin and lead, having low melting point and high resistivity.
Let a fuse wire be of length and radius r, made of a substance of resistivity .
If i be the current passing through the fuse wire, then the rate of heat generation will be
2 2
The heat dissipated to the surrounding varies directly as the surface area of the wire. If P be the thermal power generated per unit area, then the total rate of heat loss is
dE P (2 r ) dt
In the steady state the rate of heat generated due to Joule’s effect becomes equal to the rate of heat loss to the surrounding.
2
Find the heat generated in each of the resistors shown in fig. in a time interval of 1 hour. (Ignore the internal resistance of the battery)
+ – E = 24V
2
12
6
Sol. The resistances 12 and 6are connected in parallel, having an equivalent resistance of (12 ) (6 )
(12 6) 4
This 4 resistance is in series with the 2 resistance yielding an equivalent resistance of 4 + 2 = 6.
Current drawn from the battery = 24V 6 4A
Heat produced in the 2 resistance is, H = i²Rt = (4A)² (2) (3600 s) = 115.2 kJ
Since 12and 6resistors are in parallel, hence the total current of 4A gets distributed in them in the inverse ratio of their respectively resistances.
Current through the 12 and 6 resistances will be respectively.
4A 6 1.33 A
Example 33 :
Two bulbs A and B each with a rated voltage of 220V, have rated powers of 25W and 100W respectively. They are connected in series across a voltage supply of 440V. Find, which of the two bulbs will fuse.
Sol. The ratio of resistances of the bulbs A and B is,
A r B
B r A
R (P ) 100 4 R (P ) 25 1
Since, the two bulbs are in series, so the voltage across the two will be in direct proportion to their respective resistances.
A
The voltage across the bulb A exceeds the rated voltage and hence it will fuse.
Example 34 :
How much (in present) has a filament diameter decreased due to evaporation if the maintenance of the previous temperature required an increase in voltage by 0.50% ?
Sol. From eq.
, ignoring any change in dimensions, due to temperature rise,
2 k
V r where k is a positive constant.
Taking log of both sides yields, 2 ln V = ln k – ln r Differentiating, both sides yields 2dV dr
V r
Thus, the radius (and hence the diameter) has decreased by 1.0%
CHEMICAL EFFECT OF CURRENT