CURRENT ELECTRICITY
(Important formulae and Concepts)
HEATING EFFECT OF CURRENT
QUESTIONS FOR SUPPORTIVE LEARNERS
1. Sketch the graph showing the variation of resistivity of carbon with
temperature.
2. Write the mathematical relation between mobility and drift velocity of
charge carriers in a conductor. Name the mobile charge carriers
responsible for conduction of electric current.
2. The variation of potential difference V with length l in case of two
potentiometers P and Q is as shown. Which one of these two will you
prefer for comparing emf’s of two primary cells?
3. You are given ‘n’ resistors , each of resistance ‘r’ . These are first
connected to get minimum possible resistance. In the second case, these
again connected differently to get maximum possible resistance.
Compute the ratio between the minimum and maximum values of
resistances obtained.
Draw a circuit diagram using a meter
bridge and write the necessary mathematical relation used to determine
the value of an unknown resistance. Why can not such an arrangement be
used for measuring very low resistance.
4. Two cells E1 and E2 in the given circuit diagram have an emf of 5V and
9V and internal resistance of 0.3 and 1.2 respectively. Calculate the
value of current flowing through the resistance of 3.
5. Two metallic wires of the same material have the same length but crosssectional area is in the ratio 1:2. They are connected (i) in series and (ii)
parallel. Compare the drift velocities of electrons in the two wires in both
the cases (i) and (ii).
6. (i)Calculate the equivalent resistances of the given electrical network
between the points A and B.
(ii) Also calculate the current through CD and ACB , if a 10 V d.c. source
is connected between A and B, and the value of R is assumed as 2.
7. A cylindrical metallic wire is stretched to increase its length by 5%.
Calculate the percentage change in its resistance.
8. Write the mathematical relation for the resistivity of a material in terms
of relaxation time,number density and mass and charge carriers in it.
Explain, using the relation, why the resisitivity of a metal increases and
that of a semiconductor decreases with rise in temperature.
9. For the potentiometer circuit shown in the given figure, points X and Y
represents the two terminals of an unknown emf E’ . A student observed
that hen the jockey is moved from the end A to the end B of the
potentiometer wire, the deflection in the galvanometer remains in the
same direction. What may be the two possible faults in the circuit that
could result in this observation? If the galvanometer deflection at the end
B is (i) is more (ii) less, than that at the end A, which of the two faults,
listed above, would be there in the circuit? Give reason in support of your
answer in each case.
A. (a)Two possible faults are (i) the emf applied across AB is less than the
unknown emf. (ii) –ve terminal of the source of unknown emf is joined
with end A of the wire.
(b) The galvanometer deflections at the end B is more means source of
unknown emf have beeen joined with its negative terminal to end A.
Current gets divided at point A and combines at point B.The
galvanometer deflection at the end B is less than at the end A means
theemf applied is less than the unknown emf used. Currents get combined
at end A and divided at end B.
10.A 10 m long wire of uniform cross-section of 20 resistance is fitted on
the board. The wire is connected in series with a battery of 5V along
with an external resistance of 480. If an unknown e.m.f. E is balanced
at 600 cm of this wire, calculate (i) the potential gradient of wire, (ii)
value of unknown e.m.f.
H O T S ( QUESTIONS FOR HIGH ACHIEVERS)
1. Two wires of aluminium and copper have the same resistance and same length.
Which of the two is lighter? Density of copper is 8.9 × 103 kg/m3 and that of
aluminium is 2.7 × 103 kg/m3. Theresistivity of copper is 1.72 × 10–8 Ω m and that of
aluminium is 2.6 × 10–8 Ω m.
Ans. That wire will be lighter which has less mass. Let suffix 1 represent aluminium
and suffix 2represent copper.
2. Determine the current drawn by a 12V battery with internal resistance 0.5by the
following infinite network.
Ans. Let x be the equivalent resistance of the network. Since the network is infinite,
the addition of one set of three resistances, each of 1will not change the total
resistance, i.e., it will remain x. The network would then become as shown in Fig.
6.35. The resistances x and 1are in parallel and their total resistance is RP given by
3. 100 cells are arranged in rows of 50 in series and two rows in parallel. The battery
supplies a current of 1A to an electric lamp having a resistance of 50 . If the internal
resistance of each cell is 2, find the e.m.f. of each cell.
Ans. Let E volts be the e.m.f. of each cell.
4. Find the equivalent resistance between A and B
The circuit shown in Fig. (i) is equivalent to the circuit shown in Fig (ii). Resistance
of upper branch in Fig. (ii)
5. A galvanometer together with an unknown resistance in series is connected across
two identical batteries of each 1.5 V. When the batteries are connected in series, the
galvanometer records a current of 1 A and when the batteries are connected in parallel,
the current is 0.6 A. What is the internal resistance of each battery?
Ans.
6. 12 cells each having the same e.m.f. are connected in series and are kept in a closed
box. Some of the cells are wrongly connected. The battery is connected in series with
an ammeter and two cells identical with the others. The current is 3 A when the cells
and battery aid each other and 2 A when the cells and battery oppose each other. How
many cells in the battery are wrongly connected ?
Ans.
7. Calculate the steady state current in 2 resistor in the circuit shown in Fig. The
internal resistance of the battery is negligible and the capacitance of the condenser.
Ans.
8. Fig. shows a potentiometer circuit for comparing two resistances. The balance point
with a standard resistance R = 10.0is found to be at 58.3 cm while that with the
unknown resistance X is at 68.5 cm. Determine the value of X. What would you do if
you fail to find a balance point with the given cell?
Ans. We can find the value of low resistances by potentiometer. In this method,
instead of comparing the two e.m.fs, potential drops across an unknown resistance X
and a standard resistance R are compared. Referring to Fig, suppose the cell E sends a
current I through R and X. The p.d. across the standard resistance R (V1 = IR) is
balanced against a length l1 (= 58.3 cm) of the potentiometer wire. Similarly, p.d.
across unknown resistance X (V2 = IX) is balanced against a length l2 (= 68.5 cm) of
the potentiometer wire.
.
If there is no balance point, it means that the potential drop across R or X is greater
than the p.d. across the potentiometer wire AB. The obvious remedy is to reduce
current through R–X circuit. This can be achieved by placing an external resistance in
series with R and X or by using a cell of smaller e.m.f. Another remedy is to use a
battery of higher e.m.f. that sends current through the potentiometer wire
9.The total length of a potentiometer wire is 10m. The distance between the null
pointson the potentiometer wire for two cells is 60 cm. If the difference between the
e.m.fs of the cells be 0.4V,calculate (i) the potential gradient along the wire, (ii) p.d.
between the ends of this wire
10. A generator supplies power to a factory through cables of total resistance 20 ohms.
The potential difference at the generator is 5000 V and power output is 50kW.
Calculate (i) power supplied by the generator, (ii) potential difference at the factory
Ans.
11.A battery of internal resistance 4is connected to the network of resistances as
shown in Fig. What must be the value of R so that maximum power is delivered to the
network? What is the maximum power ?
. The circuit shown in Fig. is equivalent to the circuit shown in Fig. 8.13. Referring to
Fig. it is a balanced Wheatstone bridge because the products of opposite arms of the
bridge are equal.Therefore, points B and D are at the same potential. As a result, no
current flows through 6R resistance and may be considered as removed. In the branch
ADC, the total resistance = 2R + 4R = 6R and in the branch ABC,
total resistance = R + 2R = 3R.
EDITED BY:
Mr. S.K.BHAT (VICE PRINCIPAL)
Mr. A.K.SHARMA(P.G.T)