QUESTIONS ON OHM’S LAW ANDKIRCHHOFF’S LAW

A - EASY TYPE01What is drift velocity? Define mobility.02.Define the term ‘resistivity’. Give its SIunit.03.Define temperature coefficient ofresistance. What is super conductivity?04.Define ‘critical temperature’ and ‘criticalfield’ for a super conductor.05.What are thermistors? Mention its uses.

• 06.Define e.m.f. and internal resistance ofa cell.

• 07.State Kirchhoff’s law on electricalnetwork.

• 08.State Ohm’s law and mention one of itslimitations. 09.What is the condition for thebalanced state of a wheat stone’snetwork?

• 10.Give the SI unit of mobility• 11.What is a node? What is an electrical

loop?

B - APPLICATION TYPE01.How does resistance of a i) conductorii) thermistor vary with the increase in

temperature?02.What is the colour of the 2nd and 3rd

band in a coded resistor ( 5.4 X 103 )Ω?03.When current is drawn from a battery,

terminal P.D. is less than e.m.f. Why?04.What happens to the balanced state of a

wheat stone’s network when cell and thegalvanometer are interchangedgalvanometer is replaced by anothergalvanometer of different resistance.

05.A resistor of resistance R is drawn sothat its length is doubled. What is it’snew resistance.

06.How is resistance of a conductor relatedto i) lengthii) cross sectional area.

07.Mention the principle behind Kirchhoff’s1st and 2nd law

1. Using Ohm’s law, derive the expressionfor the electric current through anexternal resistor when it is connected to asource of e.m.f.

2. Derive the expression for branch currentwhen two resistor are in parallel.

3. Using Kirchhoff’s law, arrive at thecondition for the balanced state of awheat stone’s network.

C - QUESTIONS FOR LONG ANSWERS1. Define ‘strength of the electric current’.

Derive a relation between electric currentand drift velocity.

2. Assuming the expression for electriccurrent in terms of drift velocity, deduceOhm’s law. Define SI unit of resistance.

3. What is effective resistance? Derive theexpression for effective resistance whenthree resistors are in series/parallel

NOTE1.Drift velocity is the average velocity with

which the free electrons are drifting underthe applied electric field in a conductor. Asthe temperature increases the collisionbecomes more frequent than earlier. i.e.average relaxation time decreases andresistivity increases and hence resistanceof a conductor increases.

2.Drift velocity per unit electric field appliedis called mobility.Strength of the electriccurrent is directly proportional to driftvelocity.

3. Resistance is the opposition offered bythe conductor for the flow of charge.

4.Resistivity of the material of the conductoris the resistance offered by the conductorof unit length and unit cross-sectionalarea.SI unit: Ohm meter

5. Resistivity of the material of theconductor increases as the temperatureincreases.

• 06.The ratio of change in resistance per 0Crise in temperature and its original resistanceat 00C is called temperature coefficient ofresistance. If the value of is negative, thismeans as the temperature increases,resistance decreases.

• 07.Critical temperature is the temperature atwhich the resistance of the given materialabruptly falls to zero. Critical magnetic field isthe minimum magnetic field at which a superconductor returns to its normal state whenthe temperature is kept constant (Appliedexternal magnetic field has an adverse effecton super conductivity.)

• Let L be the length of the conductor, A bethe cross-sectional area of the conductor.Let V be the P.D. across the ends of theconductor, n be the number conductionelectrons present in the conductor per unitvolume. Let e be the charge and m be themass of the electron. Let Vd be the driftvelocity of the free electron and t be thetime taken by the electron to travel thelength L. We know, total charge of the freeelectron =q =n A L e.

• AnddnAeVA

tnALe

tqI ===

tLwhere Vd=

,

We know,I = n e A Vd

-----------(1)

butVd = a T´

• where T = average relaxation time

mEe

mFa == d

EeTV m\ =---------- (2)

Sub (2) in (1) we get

2ne AETI m= -----------(3)

butLVE = numerically

2 2.ne AVT ne TAI VmL mL\ = =

2mL LV I I RIAne TA

r= = × =

RVI =\ aOR I V

where

is a constant calledresistivity of the material of theconductor

2m

ne Tr =

•• is a constant called resistance of

the conductor

ALR r=

Let R1, R2, R3 be the resistances ofthree resistors connected in series.Let V1, V2, V3 be the P.D. across themrespectively. Since they are in series,the same current I flows through each.

• For series combination of resistors,• P.D. between the ends of the

combination is equal to the sum of theP.D. across the individual resistors.

\

V= V1 +V2 +V3 But, V1 =IR1V2 = IR2V3 = IR3

\

V = I (R1 + R2 + R3) ---------------(1)

• For the effective resistor Rs, the P.D. is Vand the current flowing is I.

• V = I Rs -----------------------------(2)

From (1) and (2) we have,I Rs = I (R1 +R2 +R3)

OR Rs = R1 +R2 +R3

Let R1, R2, R3 be the resistances of three resistorsconnected in parallel. Let I be the current drawn from thecell.Let I1, I2, I3 be the currents flowing throughR1, R2, R3 respectively. Since the resistors are inparallel, the P.D. across each resistor is same.

• For parallel combination of resistors,• the current drawn from the cell is equal

to the sum of the currents flowingthrough individual resistor.

1RV

2RV

3RV

\I = I1 + I2 + I3 But, I1 =

I2 =

I3 =

•

÷÷ø

öççè

æ++=

321 R:1

R1

R1VI ------(1)

For the effective resistance RP ,the P.D. is V and the current is I

PRVI = ---------(2)

• From (1) and (2) we have,•

=PR

V1 2 3

1 1 1VR R R

æ ö+ +ç ÷

è ø

PR1

321 R1

R1

R1

++=

• Common mistakes in the abovederivations

• Use of the word ‘potential difference at apoint instead of potential differencebetween two points.

• Use of the word current across theconductor instead of current flowingthrough the conductor.

• No mention of the direction of the current.• Students may also notice that effective

resistance is the resistance of the singleresistor which produces the same effect asproduced by the combination ofresistances.

Let R be the resistance of the external resistorconnected to a cell of e.m.f E,internalresistance ‘r’. Let I be the strength of the currentflowing through R. Let V be the terminal P.D.We know, work done in shifting unit positivecharge once round the closed circuit is e.m.f.of the cell.

• Hence e.m.f. is sum of work done inshifting unit positive charge oncethrough R and r.

• But, work done in shifting unit positivecharge through R is P.D. across Rand through r is P.D. across r.

• Let V and V1 be the P.D. across Rand r respectively.

• E =V + V1 (from the principle ofconservation of energy)

• From Ohm’s law, we have• V = I R and V1 = I r• E = I R + I r = I (R + r)• OR

• I =rR

E+

Note: ( i) V= I R= rRER+

is the terminal P.D.

•

• Hence, E = V when r = 0 or R = ¥Rr1

EV+

=

i.e. e.m.f. of a cell is measured by the P.D.across the cell when the circuit is open ie R = ¥(ii) A part of the e m f is used to drive the chargethrough internal resistance. Since r is notequal to zero for any cell, e.m.f. is greaterthan terminal P.D. when current is drawn from the cell

• Generally noticed mistakes:• Without explaining why E = V + V¢,

students substitute V = I R, V¢ = I r.• No mention of the value of the current

and its direction

1st LAW: Algebric sum of the electriccurrents meeting at a node is zero

I1 – I2 – I3 = 0

Students generally draw largenumber of currents and write thesign of the currents wrongly inthe above equation.

2ND LAW: In a closedelectric network, thealgebraic sum of the e.m.f.is equal to algebraic sum ofthe IR products in thatnetwork.

In the network ABCAI1 R1 + I2 R2 - I3 R3 = - E1 + E2

• General mistakes done by thestudents.

• All IR products and e.m.f.s aretaken positive without knowingwhen are they negative.

• Without mentioning the network IRproducts and e.m.f. s are equated.

• PRECAUTIONS:• While solving the problems on

Kirchhoff’s Law, it is better to retain thesolved values of the currents I1 & I2 inthe fraction form. This helps to checkthe correctness of the current values.

• In a given problem, if the cells are ofdifferent e.m.f.s and also in parallel, thatproblem can be solved using onlyKirchhoff’s Law.

• While solving the problems on Ohm’slaw a) if internal resistance is there for agiven cell, use the equation

E

e.m.f. in the circuitI = R +r

E

VR

where RE = effective external resistanceb) If terminal P.D. is given,

use I =

• P, Q, S and R are the four resistancesconnected in cyclic order to form aWheatstone’s network ACDB as shown.Let I1, I 2, I 3, I 4 and I g be the currentsflowing through P, R, Q, S and thegalvanometer of resistances Grespectively in the directions as shown.

• From Kirchhoff’s 1st law we have,• I1 = Ig+I3 at the node C – (1) and• I2+Ig = I4 at the node D – (2)

• From Kirchhoff’s 2nd law we have,• I1P + IgG – I2 R = 0 for the mesh ACDA,• --------- (3)• I3Q – I4 S – Ig G = 0 for the mesh CBDC• -----------(4)• For the balanced state of the

Wheatstone’s network, the currentflowing through the galvanometer mustbe zero

• i.e. Ig = 0

• I1 = I3 from (1) and I2 = I4 from (2)• I1 P = I3 R from (3) and I2 Q = I4 S from (4)• gives

• OR

( )( )43

SIRI

QIPI

4

2

3

1 =

SR

QP=

• Note:• To verify the balanced state of the

given Wheatstone’s network, matchthe product of 1st and 3rdresistances with the product of 2ndand the 4th resistances when fourresistances are given in cyclic order.

• Cell and the galvanometer can beinterchanged without effecting thebalanced condition of the network.