CHAP # 14EXERCISEANS no1The rate of flow of electric charges through any cross-sectional area of conductor is called electric current.It is denoted by IIts formula is

I=q/t

ANS no2

ELECTRONIC CONVENTIONALElectronic current flows from low to high potential.

Conventional current flows from high to low potential.

Electronic current flows due to negative charge carriers

conventional current flow due to positive charge carriers

Electronic current is actual current

Conventional current is just imaginary current flowing opposite to electronic current.

ANS no3e.m.f stands for electromotive force. This is not force but energy require to move charges from one point to another. ANS no4

e.m.f is the total amount of energy supplied by battery or the cell in moving a charge between two terminals of battery.The current passes through resistor and use potential difference to do usefull work.e.m.f is the energy by which other forms of energies are being converted into electrical energy.Potential difference convert electrical energy into other forms of energy.ANS no5WHOLE TOPIC OF OHM’S LAWANS no6WHOLE TOPIC OF RESISTANCE (GIVEN AFTER OHM’S LAW)ANS no7

CONDUCTORS INSULATORSThey conduct electricity. They don’t conduct

electricity.They have free electrons.

They don’t have free electrons.

Their resistance is low. Their resistance could be infinity.

ANS no8WHOLE TOPIC OF JOULE’S LAW.ANS no9

D.C A.CStands for Direct Current.

Stands for Alternating Current.

Polarity remains same.

Polarity reverse after specific time intervals.

Dangerous. Safe.For example cells. For example A.C generator.

ANS no10WHOLE TOPIC OF PARALLEL COMBINATION OF RESISTORS.ANS no11

WHOLE TOPIC OF SERIES COMBINATION OF RESISTORS.

ANS no12WHOLE TOPIC OF HAZARDS OF ELECTRICITY.ANS no13SAFETY DEVICES (LAST THREE TOPICS OF FUSE, CIRCUIT BREAKER, AND GROUND WIRE).ANS no14 a)

b)

c)All appliances get same voltage and if any one

appliance fuses then it has no effect on working of other appliances.CONCEPTUALANS no1 Conductors are the materials having large number of free electrons. These electrons could move randomly through spacing between positive charges. Because these positive charges are vibrating in a fixed position. So only electrons could conduct electricity.ANS no2 Both these are source of e.m.f. Cell is such source in which only chemical energy is converted into electrical energy. While in battery any kind of energy could be converted into electrical energy.ANS no3 All conductors consist of large number of electrons moving randomly due to which net current through conductor without battery is 0. But when battery is attached then these electrons start moving in a specific direction making current. So it is impossible to get electrical current without P.D.ANS no4

If these points are on conductor then charges would flow giving current. But if these points are on insulator then it does not constitute electricity. So if there is potential difference between two points then it is not necessary that charge will flow between these two points.ANS no5 Ammeter is connected in series because in series combination current remain same. And same current will flow through ammeter as that in circuit.

ANS no6 For measuring voltage of circuit is connected in parallel with voltmeter because in parallel combination voltage remain same.So voltage of circuit will also be given to voltmeter.

ANS no7

1Wh=3600J1Wh=3600 x 1J1/3600 =1JMULTIPLYING BOTH SIDES BY 1000(1/3600) x 1000 = 1000J1000/3600 =1000J0.278Wh=1000J

ANS no8

Headlights are connected in parallel because sometimes only one light is working and other is out of order. This could only happen if they are in parallel.ANS no9 Bulb with resistance of 5Ω will give brighter lightbecause of low resistance as compared to 10Ω.And as we know P=I2RSo same bulb of 5Ω will discharge first.ANS no10 Because in series circuit if any one appliance fuses then other also stops working. Also both of these require different amount of current.ANS no11 Fuse controls the current. As it is connected in series so that when large amount of current passes through wires then it could cut the circuit by melting itself.

ANS no12 Because in one wire there is no potential difference and for the flow of current there should be potential difference and hence birds could easily sit on wire.NUMERICALS14.1

GIVENI=3mA=3x10-3

t=1min=60secREQ

Q=?SOLUTION

I=Q/tQ=I x tQ=3 x 10-3 x 60Q=180 mC

14.2GIVEN

R=100000Ω

REQI1=? (if V=12V)I2=? (if V=12V, R=1000Ω)

SOLUTIONV=IRI=V/RI1=V/RI1=12/100000I1=1.2 x 10-4AI2=V/RI2=12/1000I2=1.2 x 10-2A

14.3GIVEN

R=10MΩV=100V

REQI=?

SOLUTIONV=IR

I=V/RI=100/(10 x 106)I=1 x 10-5AI=0.01mA

14.4GIVEN

V=10VI=1.5At=2min=2x60=120sec

REQE=?

SOLUTIONW=E=I2RtE=I2(V/I)t (AS R=V/I)E=(1.52)(10/1.5)x120E=1800J

14.5GIVEN

R1=2kΩR2=8kΩV=10V

REQRe=?I1=I2=I=?V1=?V2=?

SOLUTIONRe=R1+R2

Re=2 + 8Re=10kΩI=V/Re

I=10/10000I=x 10-3AI=1mAI1=I2=1mAV1=IR1

V1=1m x 2kV1=2V

V2=IR2

V2=1m x 8kV2=8V

14.6GIVEN

R1=6kΩR2=12kΩV=6V

REQRe=?V1=V2=?I1=?I2=?

SOLUTION1/Re=1/R1 + 1/R2

1/Re=1/6 + 1/121/Re=1/4Re=4kΩAS IN PARALLEL COMBINATION VOLTAGE

REMAIN SAME SO V1=V2=6V

I1=V/R1

I1=6/6I1=1mAI2=V/R2

I2=6/12I2=0.5mA

14.7GIVEN

V=220VP=100Wt=5hrstmonth=5 x days of monthtmonth=5 x30tmonth=150 hr

REQE(kwh)=?

SOLUTIONP=V2/RR=V2/PR=(2202)/100R=484kΩ

E(kwh)=PtE(kwh)=100 x 150E(kwh)=15000whE(kwh)=15 x 103whE(kwh)=15kwh

14.8GIVEN

R=95ΩP=150W

REQV=?

SOLUTIONP=V2/RV2=PRV2=150 x 95

SQUARE ROOTING BOTH SIDESV= 120V

14.9GIVENa)

n=no. of bulbs=10P(bulb)=60Wt=5hrP(n)=n x P(bulb)

P(n)=10 x 60P(n)=600WP(a)(kwh)=(P(n) x t)/1000P(a)(kwh)=(600 x 5 x 30)/1000P(a)(kwh)=90kwh

b)n=no. of fans=4

P(fan)=75Wt=10hrP(n)=n x P(fan)

P(n)=4 x 75P(n)=300WP(b)(kwh)=(P(n) x t)/1000P(b)(kwh)=(300 x 10 x 30)/1000P(b)(kwh)=90kwh

c)n=no. of T.V=1

P(T.V)=250Wt=2hrP(n)=n x P(T.V)

P(n)=1 x 250P(n)=250WP(c)(kwh)=(P(n) x t)/1000P(c)(kwh)=(250 x 2 x 30)/1000P(c)(kwh)=15kwh

d)n=no. of iron=1P(iron)=1000Wt=2hrP(n)=n x P(iron)

P(n)=1 x 1000P(n)=1000WP(d)(kwh)=(P(n) x t)/1000P(d)(kwh)=(1000 x 2 x 30)/1000P(d)(kwh)=60kwh

Cost of one unit=Rs.4REQ

Bill=?SOLUTION

P(t)= P(a)(kwh)+ P(b)(kwh)+ P(c)(kwh)+ P(d)(kwh)

P(t)=90+90+15+60P(t)=255kwhBill=cost of one unit x P(t)

Bill=Rs.4 x 255Bill=Rs.1020/-

14.10GIVEN

P1=100WP2=4kW=4 x 103WV=250V

REQI1=?I2=?R1=?R2=?

SOLUTIONP1=I1V

I1=P1/VI1=100/250I1=0.4AI2=P2/V

I2=(4 x 1000)/250

I2=8A

V=I1R1

R1=V/I1

R1=250/0.4

R1=625Ω

V=I2R2

R2=V/I2

R2=250/8

R2=31.25Ω

14.11GIVEN

R=5.6ΩV=3VI=0.5A

REQP(dissipated)=?P(T)=?

REASON FOR DIFFERENCE OF P(dissipated) AND P(T)=?SOLUTION

P(dissipated)=I2RP(dissipated)=0.52 x 5.6P(dissipated)=1.4WP(T)=VIP(T)=3 x 0.5P(T)=1.5W

REASON FOR DIFFERENCE OF P(dissipated) AND P(T)= 0.1W is lost due to internal resistance of the battery.