The document provides the solution to a physics problem involving two resistors connected in series and parallel configurations. When connected in series, their effective resistance is 3 ohms, and when connected in parallel it is 2/3 ohms. Using these values and equations for series and parallel resistances, the problem determines that the individual resistances of the two resistors are 2 ohms and 1 ohm.

1. CURRENT ELECTRICITY
QUESTION BANK (3 MARKS)
1
A
Two resistors when connected in series their effective resistance is 3 Ω and when
connected in parallel their effective resistance is
2
3
Ω. Then find the value of each
resistance.
Let R1 andR2 be the resistancesof the resistors.
AffectiveResistance whenconnectedinseries= R1 + R2 = 3Ω-------------(1)
AffectiveResistance whenconnectedinparallel=
R1R2
R1+ R2
=
2
3
Ω
R1R2
3
=
2
3
Ω i.e., R1R2 = 2Ω
(R1 − R2)2 = (R1 + R2)2 − 4R1R2 = 32 − 4X2 = 9 − 8 = 1
R1 − R2 = 1---------------------------------------------(2)
Solvingequations(1) and(2) R1 = 2Ω and R2 = 1Ω
1/
2
1/
2
1
1
2
A
Give the graphical relationship between the potential difference and current for a
conductor obeying Ohm’s Law. State the conditions under which the Ohm’s Law
holds good. How will you find the resistance of the conductor from the graph?
Graphical relationshipbetweenthe potential differenceandcurrentfora conductor
obeyingOhm’slawis linear.
v
I
Conditions under which the Ohm’s Law holds good are
Temperature iskeptconstant
Resistance of the conductorcan be foundout byfindingthe slope of the line
𝑅 =
∆𝑉
∆𝐼
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 𝑅 =
𝑉2
𝑃
=
100𝑋100
100
= 100Ω
For 40 W bulb 𝑅 =
𝑉2
𝑃
=
100𝑋100
40
= 250Ω
The total resistance inthe circuitis 100 + 250 = 350Ω
The current flowingthroughthe circuitis 𝐼 =
𝑉
𝑅
=
100
350
= 0.29𝐴𝑚𝑝
For the 100 W bulb 𝑃 = 𝐼2 𝑅 = 0.292 𝑋100 = 8.41𝑊
For the 40W bulb 𝑃 = 𝐼2 𝑅 = 0.292 𝑋250 = 21𝑊
The more power the brighter the bulb.Hence 40 W bulbglowsbrighter than 100W
bulb.
½
½
½
½
½
½
4 Draw the circuit diagram used to measure the current through a material obeying
Ohm’s 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 Ohm’s Law when the
Voltage is applied across the material.
Formulato findthe resistance of the material R=
𝜌𝑙
𝐴
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 𝑅 =
𝜌𝑙
𝐴
Where 𝜌 is Resistivity of the material.
a. Since resistance doubles.
b. Since 𝑅 ∝
1
𝐴
𝑖. 𝑒. , 𝑅 ∝
1
𝑟2 where A = 𝜋𝑟2
= 𝜋(
𝐷
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 ‘O’ 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.
𝑉 =
𝑊
𝑄
, where W is the work done and Q is the amount of charge moved
𝑉 = =
20𝐽
20𝑋 10−3 = 103
𝑉
Current flowing through the material is given by 𝐼 =
𝑄
𝑡
=
20𝑋 10−3
20
= 10−3
𝐴
Therefore 𝐼 = 10−3
𝐴.
½
1
½
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 Ω.
In the above net work all the resistors are connected in parallel and the
value of each resistor is 2Ω.
The effective resistance is 𝑅 =
2
3
Ω
𝐼 =
𝑉
𝑅
=
10
2
3
= 𝑅 =
30
2
=15 A
½
1
½
1
10
A
a. The charge possessed by an electron is 1.6 X 10−19
𝐶. Find the
number of electrons that will flow per second to constitute a
current of 1 ampere.
b. A 9Ω 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 10−19
𝐶
Charge 1𝐶 = 𝑞 = 𝑛𝑒 = 𝑛 𝑋 1.6 X 10−19
𝐶
∴ 𝑛 =
1𝐶
1.6 X 10−19 𝐶
=
1
1.6 X 10−19 =
1019
1.6
= 6.25X 1018
Since a current of 1 amp=
1𝐶
1𝑠𝑒𝑐
that is for a current of 1 amp the number of
electrons that will flow per is 6.25X 1018
b. When a 9Ω resistor is cut into three equal parts the resistance of each
part is 𝑅 =
9
3
= 3Ω
When all the parts are connected in parallel we get network to be as
shown in the diagram
The effective resistance is
1
𝑅
=
1
𝑅1
+
1
𝑅2
+
1
𝑅3
=
1
𝑅
=
1
3
+
1
3
+
1
3
=
3
3
= 1
½
1
½
1
The value of
each resistance
is3Ω.

5. ∴ 𝑅 = 1Ω
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= 2.5𝑉 𝑋 750𝑚𝐴 = 2.5𝑋750 𝑋 10−3
= 187.5𝑊
Resistance of the torch(R )=
𝑉
𝐼
=
2.5
750 𝑋 10−3 =3.33Ω
Energy consumed by the bulb=Power X Time =187.5W X 4Hrs=750 watt hours.
1
1
1
12
Ans
Two resistors with resistance 5Ω and 10Ω 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= 5Ω and R2=10 Ω
When connected in series effective resistance R= R1 + R2= 5+10=15Ω
Current flowing through the circuit 𝐼 =
𝑉
𝑅
=
6
15
= 0.4 𝐴
Whenconnectedinparallel
1
𝑅𝑝
=
1
𝑅1
+
1
𝑅2
=
1
5
+
1
10
𝑅𝑝 =
𝑅1𝑅2
𝑅1+𝑅2
=
5 𝑋 10
5+10
=
50
15
=3.33Ω
Current through the circuit is 𝐼 =
𝑉
𝑅
=
6
3.33
1.8 Ω
Current will be minimum when the resistors are connected in series and
½
1
½
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 10−8
Ωm.
Length of the wire 𝑙 = 2𝑘𝑚 = 2000𝑚
Radius of the wire𝑟 = 2𝑚𝑚 = 2X 10−3
𝑚
Area of cross section=𝜋𝑟2
=
22
7
𝑋(2𝑋 10−3
)2
=
22
7
𝑋(4𝑋 10−6
)𝑚2
Resistivity of the wire=1.92 X 10−8
Ωm
𝑅 =
𝜌𝑙
𝐴
=
1.92 X 10−8
Ωm X 2000m
22
7
𝑋(4𝑋 10 −6)𝑚2
=3Ω ∴ 𝑅 =3Ω
½
½
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 Ω connected in parallel across a supply voltage of 10V.Also find the
current flowing through each resistor.

7. Ans
𝑅1 = 𝑅2 = 𝑅3 = 3Ω Supply voltage is 10V
Net resistance
1
𝑅
=
1
𝑅1
+
1
𝑅2
+
1
𝑅3
=
1
𝑅
=
1
3
+
1
3
+
1
3
=
3
3
= 1
∴𝑅=1Ω
Current flowing through the circuit= 𝐼 =
𝑉
𝑅
=
10
1
= 10𝐴𝑚𝑝
Current flowing through each resistor is same and is equal to
10
3
=
3.33𝐴𝑚𝑝
17 Find the current flowing through a series combination of resistors 5Ω, 10Ω,3Ω
connected across a voltage of 36V. Also find voltage across each resistor.
𝑅1 = 5Ω 𝑅2 = 10Ω 𝑅3 = 30Ω
𝑅 = 𝑅1 + 𝑅2 + 𝑅3=5+10+30= 18 Ω
Current flowing through the circuit =𝐼 =
𝑉
𝑅
=
36𝑉
18Ω
= 2𝐴𝑚𝑝
Voltage across each resistor
𝑉1 = 𝐼𝑅1 = 2𝑋5 = 10 V
𝑉2 = 𝐼𝑅2 = 2𝑋10 = 20 V
𝑉3 = 𝐼𝑅3 = 2𝑋30 = 60 V
½
½
½
½
½
½
18
Ans
A metallic wire of length ‘l’ and area of cross section ‘A’ has a resistance “R”. 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 Ω
5Ω
10Ω36V

8. V= 𝑙1𝐴1 = 𝑙2𝐴2 ;
𝑙1
𝑙2
=
𝐴2
𝐴1
Whenthe wire isstretchedtodouble itslength 𝑙2 = 2𝑙1; 𝐴2 =
𝐴1
2
Before stretchingresistance of the wire is 𝑅1 =
𝜌𝑙1
𝐴1
Afterstretchingresistanceof the wire is 𝑅2 =
𝜌𝑙2
𝐴2
=ρ
2𝑙1
𝐴1
2
=𝜌4
𝑙1
𝐴1
(ρ doesnot change for the same material)
Hence 𝑅2 =4ρ
𝑙1
𝐴1
= 4 𝑅1 Resistance increases
b. Resistivity
Resistivitydoesnotchange forthe same material
½
½
½
½
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
∆V
∆I
Current(Amp)
V
o
l
t
a
g
e
(
v
o
l
t
s
)
Slope of the line
R=
∆𝑉
∆𝐼
=
4.5−2.5
0.8−0.4
=
2
0.4
= 5Ω

9. and is equal to 5Ω.
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 𝑅 =
𝑉2
𝑃
=
220𝑋2200
100
= 484Ω
For 40 W bulb 𝑅 =
𝑉2
𝑃
=
220𝑋220
60
=
2420
3
Ω
The current drawn by the bulb of resistance 484 Ω i.e., 100W bulb is
𝐼1 =
𝑉
𝑅1
=
220
484
= 0.45𝐴𝑚𝑝
The current drawn by the bulb of resistance
2420
3
i.e., 60W bulb is
𝐼2 =
𝑉
𝑅2
=
220
2420
3
= 0.27𝐴𝑚𝑝
½
½
1
1