(b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.
In series, total resistance Rs = R1 + R2+ R3 = 1+2+3= 6 Ώ
(b) Current through he circuit , I = V/ Rs = 12/6 = 2 A
∴ Potential drop across R1 = I R1 = 2×1 = 2 V
Potential drop across R2 = I R2 = 2×2 =4V
Potential drop across R3 = I R3 = 2×3=6V
(b) If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.
In parallel combination, total resistance Rp is given by
(b) Current through R1 = V/ R1 = 20/2 = 10 A ; Current through R2= 20/4 = 5 A
Current through R3 = 20/5 = 4 A
According to Kirchhoff’s second law;
In a closed circuit EABCE ;
1 10 = 10(i1-i2)+ 10 i1 + 5 (i1 – i3)
Or 10 = 25 i1 +10 i2 – 5 i3
Or 2 = 5 i1 + 2 i2 – i3
In a closed circuit BCDA ;
10 i1 + 5 i3 – 5 i2 = 0
Or 2 i1 + i3 – i2 = 0
Or i2 = 2 i1 + i3
In a closed circuit BCDB ;
5 (i1 – i3) – 10 (i1 + i3 ) – 5 i3 = 0
Or 5 i1 – 10 i2 – 20 i3 = 0 or i1 = 2i2 + 4 i3 ….(iii)
Form (ii) and (iii) ;
I1 = 2(2 i1 + i3) + 4 i3 = 4 i1 + 6 i3 or 3 i1 = – 6 i3 or i1 = – 2 i3 …..(iv)
Putting this value in (ii) ; i2 = 2 (-2 i3) + i3 = -3 i3
Putting values in (i) ; 2 = 5(-2 i3) + 2 (-3 i3) – i3 or 2 = -17 i3 or i3 = -2/17 A
From (iv) ; i1 = -2 (-2/17)= 4/17 A ;
From (v) ; i2 = – 3 (- 2/17) = (6/17) A
i1 + i2 = (4/17) + (6/17) = (10/17) A
I1 – i3 = 4/17 – (-2/17) = (6/17) A
I2 + i3 = 6/17 + (- 2 /17) = 4/17 A
(b) Determine the balance point of the above bridge if X and Y are interchanged ?
(c) What happens if the galvanometer and cell are interchanged at the balance point of the bridge ?
Would the galvanometer show any current ?
(b) As X and Y are interchanged, therefore, l1 and l2 (i.e., lengths) are also interchanged.
Hence, I = 100 – 39.5 = 60.5 cm
(c) The position of balance point remains unchanged. The galvanometer will show no current.
Voltage of d. c. supply = 120 V
Internal resistance battery, r = 0.5 Ω ;
External resistance, R = 15.5 Ω
Since a storage battery of e. m. f. 8 V is charged with a d. c. supply 120 V, the effective e. m. f. in the circuit is given by
E = 120 – 8 = 112 V
Total resistance of the circuit = R + r = 15.5 + 0.5 = 16.0 Ω
∴ Current in the circuit during charging is given by,
V = 120 – 108.5 = 11.5 V
The series resistor limits the current drawn from the external source of d. c. supply. In its absence the current will be dangerously high.
(b) A secondary cell after long use has an emf of 1.9 V and a large internal resistance of 380 Ω. What maximum current can be drawn from the cell ? Could the cell drive the starting motor of a car ?
This amount of current cannot start a car because to start the motor, the current required is 100 A for few seconds.
Relative density d1 = 2.7
Let A1, A2 be the area of cross section for aluminum wire and copper wire.
It shows that copper wire is 2.16 time =s heavier than aluminium wire. Since the same value of length and resistance, aluminium were has lesser mass then copper wire, therefore aluminium wire is preferred of overhead power cables. A heavy cable may sag down owing to its own weight.
Q.17. What conclusion can you draw from the following observations on a resistor made of alloy manganin :
|Current A||Voltage V||Current A||Voltage V|
(b) Is Ohm’s law universally applicable for all conducting elements ? If not, give examples of elements which do not obey ohm’s law.
(c ) A low voltage supply from which one needs high currents must have low internal resistance. Why?
(d) A high tension (HT) supply of say 6 kV must have a very large internal resistance. Why?
(b) Ohm’s law is not applicable for non ohmic elements. For example ; vacuum tubes, semi- conducting diode, liquid electrolyte etc.
(c) As. L max =e. m. f . /internal resistance, so for maximum current, internal resistance should be least.
(d) A high tension supply must have a large internal resistance otherwise, if accidently the circuit is shorted, the current drawn will exceed safety limit and will cause damage to circuit.
(a) Alloys of metals usually have (greater/less) resistivity than that constituent metals.
(b) Alloys usually have much (lower/ higher) temperature coefficients of resistance then pure metals.
(c) The resistivity of the alloy manganin in (nearly independent of/ increases rapidly) with increase of temperature.
(d) The resistivity of a typical insulator (e.g. amber) is greater then that of a metal by a factor of the order of (1022 or 103).
(b) Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will you combine them to get an equivalent resistance of (i) (11/3) Ω (ii) (11/5) Ω (iii) 6 Ω (iv) (6/11) Ω ?
(c) Determine the equivalent resistance of networks shown in and (b).
For minimum effective resistance, the n resistors must be connected in parallel.
Maximum effective resistance, Rp = R/n
(b) It is to be noted that (a) the effective resistance of parallel combination of resistors is less then the individual resistance and (b) the effective resistance of series combination of resistors is more than individual resistance.
Case (i) Parallel combination of 1 Ω and 2 Ω is connected in series with 3 Ω.
Effective resistance of 1 Ω and 2 Ω in parallel will be given by Rp
Case (iii) All the resistances are to be connected in series. Now
Case (iv) All the resistance are to be connected in parallel
(c) For the given net work is a series combination of 4 equal units. Each unit has 4 resistances in which, two resistance (1 Ω each in series) are in parallel with two other resistances (2 Ω each in series).
For the five resistances each of value R, are connected in series.
Their effective resistance = 5 R
Therefore, the network would appear as shown in and its total resistance should remain x. Here the parallel combination of x and 1 Ω is in series with the two resistors of 1 Ω each. The resistance of parallel combination is
The vale of resistance can be negative, therefore, the resistance of net work
(a) What is the value of E ?
(b) What purpose does the high resistance of 600 k Ω have ?
(c) Is the balance point affected by the high resistance ?
(d) Is the balance point affected by the internal resistance of the driver cell ?
(e) Would the method work in the absolute situation, if the driver cell of the potentiometer had an emf of 1-0 V instead of 2.0 V ?
(f) Would the circuit work well for determining extremely small emf, say of the order of a few mV (such as the typical emf of a thermocouple) ? If not, how will you modify the circuit ?
(b) The purpose of using high resistance of 600 k Ω is to allow very small current through the galvanometer when the movable contact is far from the balance point.
(c) No, the balance point is not affected by the presence of this resistance.
(d) No, the balance point is not affected by the internal resistance of the driver cell.
(e) No, the method will not work as the balance point will not be obtained on the potentiometer wire if the e. m. f. of the driver cell is less than the e. m. f. of the other cell.
(f) The circuit will not work for measuring extremely small e. m. f. because in that case, the balance point will be just close to the end A. To modify the circuit, we have to use a suitable high resistance in series with the cell of 2.0 V. This would decrease the current in the potentiometer were. Therefore, potential difference/cm of wire will decrease. Then only extremely small e. m. f. can be measured.
Let / be the current in the potentiometer wire and ε_1 and ε_2 be the potential drops across R and X respectively . When connected in circuit by closing respective key. Then
If there is no balance point with given cell of e. m. f. it means potential drop across R or X is greater than the potential drop across the potentiometer wire AB. In order to obtain the balance point, the potential drops across R and X are to be reduced, which is possible by reducing the current in R and X. For that, either a suitable resistance should be put in series with R and X or a cell of smaller e. m. f. E should be used. Another possible way is to increase the potential drop across the potentiometer wire by increasing the voltage of driver cell.