1. CURRENT ELECTRICITY
QUESTION BANK (3 MARKS)
1
A
Two resistors when connected in series their effective resistance is 3 β¦ and when
connected in parallel their effective resistance is
2
3
β¦. Then find the value of each
resistance.
Let R1 andR2 be the resistancesof the resistors.
AffectiveResistance whenconnectedinseries= R1 + R2 = 3β¦-------------(1)
AffectiveResistance whenconnectedinparallel=
R1R2
R1+ R2
=
2
3
β¦
R1R2
3
=
2
3
β¦ i.e., R1R2 = 2β¦
(R1 β R2)2 = (R1 + R2)2 β 4R1R2 = 32 β 4X2 = 9 β 8 = 1
R1 β R2 = 1---------------------------------------------(2)
Solvingequations(1) and(2) R1 = 2β¦ and R2 = 1β¦
1/
2
1/
2
1
1
2
A
Give the graphical relationship between the potential difference and current for a
conductor obeying Ohmβs Law. State the conditions under which the Ohmβs Law
holds good. How will you find the resistance of the conductor from the graph?
Graphical relationshipbetweenthe potential differenceandcurrentfora conductor
obeyingOhmβslawis linear.
v
I
Conditions under which the Ohmβs Law holds good are
Temperature iskeptconstant
Resistance of the conductorcan be foundout byfindingthe slope of the line
π =
βπ
βπΌ
1
1
1
2. 3
A
Which bulb among 100W and 40 W will glow brightly when connected in series?
Given that the voltage across the resistors is 100 V.
Use the powerequationtocalculate the resistance of the twobulbs.
For 100 W bulb π =
π2
π
=
100π100
100
= 100β¦
For 40 W bulb π =
π2
π
=
100π100
40
= 250β¦
The total resistance inthe circuitis 100 + 250 = 350β¦
The current flowingthroughthe circuitis πΌ =
π
π
=
100
350
= 0.29π΄ππ
For the 100 W bulb π = πΌ2 π = 0.292 π100 = 8.41π
For the 40W bulb π = πΌ2 π = 0.292 π250 = 21π
The more power the brighter the bulb.Hence 40 W bulbglowsbrighter than 100W
bulb.
Β½
Β½
Β½
Β½
Β½
Β½
4 Draw the circuit diagram used to measure the current through a material obeying
Ohmβs law when the voltage is applied across the material. Also give the formula
to find the resistance of the material and its SI unit of measurement.
Circuit diagram used to measure the current through a material obeying Ohmβs Law when the
Voltage is applied across the material.
Formulato findthe resistance of the material R=
ππ
π΄
The SI unitof measurementof Resistance isOhm.
1
1
1
5 Give any two differences between resistance of a conductor and resistivity.
Resistance Resistivity
1. It isthe opposition offered
by the conductorto the
flowof current
It isthe electrical resistance
perunitlengthandper unit
cross sectional areaof the
conductor.
2. SI unit is Ohm SI unit of measurement
is Ohm-m
1
1
R
40W100W
100V
3. 1
6
A
Answer the following
a. Why alloys are used for making standard resistance coils.
b. What material is used for making electrical fuses and why?
c. Which metal is used in making an electric heater coil?
a. Alloysare usedformakingstandardresistance coilsbecause theyhave avery
highvalue of resistivityand donot oxidiseimmediately.
b. The material usedforfuse elementsmustbe of low meltingpoint,low resistivity
so that itprotectsan electrical powersystem fromthe harmful effectsof over
currents.
c. Nichrome is usedin makingheatercoil because ithasrelativelyhighresistance.
1
1
1
7
A
How does the resistance of a material affect if
a. Its length is doubled
b. Its diameter is reduced to half.
c. What happens to the resistivity of the material if length of the
conductor is doubled and area of cross section is made half of the
original value?
Resistance of a material is given by π =
ππ
π΄
Where π is Resistivity of the material.
a. Since resistance doubles.
b. Since π β
1
π΄
π. π. , π β
1
π2 where A = ππ2
= π(
π·
2
)2
When Diameter is reduced to half, radius is reduced to half thereby
resistance is increased by 4 times.
c. Remains same.
1
1
1
8
A
If work is done in moving a charge of 20mc from infinity to a point βOβ in an
electric field is 20 J, then what is the electric potential at that point? Also find the
current flowing through the material in 20 seconds.
π =
π
π
, where W is the work done and Q is the amount of charge moved
π = =
20π½
20π 10β3 = 103
π
Current flowing through the material is given by πΌ =
π
π‘
=
20π 10β3
20
= 10β3
π΄
Therefore πΌ = 10β3
π΄.
Β½
1
Β½
1
4. 9
A
Find the equivalent resistance of the following network. Also find the current
through the circuit if the network is connected across 10 V. The value of each
resistance is 2 β¦.
In the above net work all the resistors are connected in parallel and the
value of each resistor is 2β¦.
The effective resistance is π =
2
3
β¦
πΌ =
π
π
=
10
2
3
= π =
30
2
=15 A
Β½
1
Β½
1
10
A
a. The charge possessed by an electron is 1.6 X 10β19
πΆ. Find the
number of electrons that will flow per second to constitute a
current of 1 ampere.
b. A 9β¦ resistor is cut into 3 equal parts and connected in parallel.
Find the net resistance of the combination.
a. The charge on an electron=1.6 X 10β19
πΆ
Charge 1πΆ = π = ππ = π π 1.6 X 10β19
πΆ
β΄ π =
1πΆ
1.6 X 10β19 πΆ
=
1
1.6 X 10β19 =
1019
1.6
= 6.25X 1018
Since a current of 1 amp=
1πΆ
1π ππ
that is for a current of 1 amp the number of
electrons that will flow per is 6.25X 1018
b. When a 9β¦ resistor is cut into three equal parts the resistance of each
part is π =
9
3
= 3β¦
When all the parts are connected in parallel we get network to be as
shown in the diagram
The effective resistance is
1
π
=
1
π 1
+
1
π 2
+
1
π 3
=
1
π
=
1
3
+
1
3
+
1
3
=
3
3
= 1
Β½
1
Β½
1
The value of
each resistance
is3β¦.
5. β΄ π = 1β¦
11
Ans
A torch bulb is rated 2.5 V and 750mA. Calculate its power, its resistance, its
energy consumed if the bulb is lighted for 4 hrs.
Voltage across the bulb=2.5V
Current passing through the bulb=750mA
Power=VI= 2.5π π 750ππ΄ = 2.5π750 π 10β3
= 187.5π
Resistance of the torch(R )=
π
πΌ
=
2.5
750 π 10β3 =3.33β¦
Energy consumed by the bulb=Power X Time =187.5W X 4Hrs=750 watt hours.
1
1
1
12
Ans
Two resistors with resistance 5β¦ and 10β¦ are connected to a battery 6V so as to
obtain minimum current and maximum current How you will connect the
resistors in each case. Calculate the strength of the total current in the circuit for
the two cases
Where R1= 5β¦ and R2=10 β¦
When connected in series effective resistance R= R1 + R2= 5+10=15β¦
Current flowing through the circuit πΌ =
π
π
=
6
15
= 0.4 π΄
Whenconnectedinparallel
1
π π
=
1
π 1
+
1
π 2
=
1
5
+
1
10
π π =
π 1π 2
π 1+π 2
=
5 π 10
5+10
=
50
15
=3.33β¦
Current through the circuit is πΌ =
π
π
=
6
3.33
1.8 β¦
Current will be minimum when the resistors are connected in series and
Β½
1
Β½
1
6. maximum amount of current flows through the circuit when they are connected
in parallel
13
A
Calculate the resistance of 2km long copper wire of radius 2mm. Resistivity of
copper is 1.92 X 10β8
β¦m.
Length of the wire π = 2ππ = 2000π
Radius of the wireπ = 2ππ = 2X 10β3
π
Area of cross section=ππ2
=
22
7
π(2π 10β3
)2
=
22
7
π(4π 10β6
)π2
Resistivity of the wire=1.92 X 10β8
β¦m
π =
ππ
π΄
=
1.92 X 10β8
β¦m X 2000m
22
7
π(4π 10 β6)π2
=3β¦ β΄ π =3β¦
Β½
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1
1
14
Ans
Give any three differences between series and parallel connection of resistors.
Series circuit Parallel circuit
1 Onlyone pathfor current ina
closedcircuit
Currentisdifferentthrougheach
resistorof the circuit(accordingtothe
resistorvalue)
2 Currentremainssame inall parts
of the circuit
Voltage remainsthe same acrossthe
each resistor.
3 Netresistance value increases Netresistance value decreases
1
1
1
15
Ans
Distinguish between conductors and insulators. On what factors resistance of a
conductor depends?
CONDUCTORS INSULATORS
1. These are the substanceswhich
allowthe electricitytopassthrough
them
These are the substances
whichresistthe flow of
electricitythroughthem
Examples Include metals,andaqueous
solutionsof salts
Examplespaper,rubber,
glassetc.
2 In a conductorthe outerelectrons
are looselybound andfreelymove
throughthe material whencurrent
isallowedpassthrough.
Most of the atomshold
theirelectronstightly.
Resistance dependsonlengthof aconductor,Areaof cross sectionandtemperature.
1
1
1
16 Find the current flowing through a parallel combination of three resistors each of
value 3 β¦ connected in parallel across a supply voltage of 10V.Also find the
current flowing through each resistor.
7. Ans
π 1 = π 2 = π 3 = 3β¦ Supply voltage is 10V
Net resistance
1
π
=
1
π 1
+
1
π 2
+
1
π 3
=
1
π
=
1
3
+
1
3
+
1
3
=
3
3
= 1
β΄π =1β¦
Current flowing through the circuit= πΌ =
π
π
=
10
1
= 10π΄ππ
Current flowing through each resistor is same and is equal to
10
3
=
3.33π΄ππ
17 Find the current flowing through a series combination of resistors 5β¦, 10β¦,3β¦
connected across a voltage of 36V. Also find voltage across each resistor.
π 1 = 5β¦ π 2 = 10β¦ π 3 = 30β¦
π = π 1 + π 2 + π 3=5+10+30= 18 β¦
Current flowing through the circuit =πΌ =
π
π
=
36π
18β¦
= 2π΄ππ
Voltage across each resistor
π1 = πΌπ 1 = 2π5 = 10 V
π2 = πΌπ 2 = 2π10 = 20 V
π3 = πΌπ 3 = 2π30 = 60 V
Β½
Β½
Β½
Β½
Β½
Β½
18
Ans
A metallic wire of length βlβ and area of cross section βAβ has a resistance βRβ. Now
the wire is stretched uniformly to double its length. What is its new
a. Resistance b. Resistivity
a. Resistance
Volume of the material remainsconstant
Volume of the wire before stretching=Volume of the wire afterstretching
10V
30 β¦
5β¦
10β¦36V
8. V= π1π΄1 = π2π΄2 ;
π1
π2
=
π΄2
π΄1
Whenthe wire isstretchedtodouble itslength π2 = 2π1; π΄2 =
π΄1
2
Before stretchingresistance of the wire is π 1 =
ππ1
π΄1
Afterstretchingresistanceof the wire is π 2 =
ππ2
π΄2
=Ο
2π1
π΄1
2
=π4
π1
π΄1
(Ο doesnot change for the same material)
Hence π 2 =4Ο
π1
π΄1
= 4 π 1 Resistance increases
b. Resistivity
Resistivitydoesnotchange forthe same material
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Β½
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1
19
Ans
Plot the I-V graph for the following data
V(volt) 0.5 1.0 1.5 2.0 2.5 3.0 4.0 5.0
I (Amp) 0.1 0.2 0.3 0.4 0.5 0.6 0.8 1.0
Can the resistance of the wire be determined from the data/graph?
S No current(Amp) Voltage(Amp)
1 0.1 0.5
2 0.2 1.0
3 0.3 1.5
4 0.4 2.0
5 0.5 2.5
6 0.6 3.0
7 0.8 4.0
8 1.0 5.0
5.0
4.5
4.0
3.5
3.0
2.5
2.0
1.5
1.0
0.5
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
Yes the resistance of the wire can be determined by finding the slope of the line
βV
βI
Current(Amp)
V
o
l
t
a
g
e
(
v
o
l
t
s
)
Slope of the line
R=
βπ
βπΌ
=
4.5β2.5
0.8β0.4
=
2
0.4
= 5β¦
9. and is equal to 5β¦.
20
Ans
.
Two electric bulbs rated 100 W, 220Volts and 60W, 220V are connected in
parallel to an electric supply of 220 V. Calculate the current drawn by the two
bulbs.
Use the powerequationtocalculate the resistance of the twobulbs.
For 100 W bulb π =
π2
π
=
220π2200
100
= 484β¦
For 40 W bulb π =
π2
π
=
220π220
60
=
2420
3
β¦
The current drawn by the bulb of resistance 484 β¦ i.e., 100W bulb is
πΌ1 =
π
π 1
=
220
484
= 0.45π΄ππ
The current drawn by the bulb of resistance
2420
3
i.e., 60W bulb is
πΌ2 =
π
π 2
=
220
2420
3
= 0.27π΄ππ
Β½
Β½
1
1