h electriccircuit solutions

8/13/2019 H ElectricCircuit Solutions

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Solutions to Electric Circuit1. A battery with an emf of 12 V and an internal resistance of 1 is used to charge a

battery with an emf of 10 V and an internal resistance of 1

. The current in thecircuit is:

A) 1 A

B) 2 A

C) A

!) 11 A

") 22 A

#olution: All elements are in series:

2 10 12I+ = 1I A=

Ans: A

2. Two 110$V light bulbs% one &2' (& and the other &100 (&% are connected in series

to a 110 V source. Then:

A) the current in the 100$( bulb is greater than that in the 2'$( bulb

B) the current in the 100$( bulb is less than that in the 2'$( bulb

C) both bulbs will light with eual brightness

!) each bulb will ha*e a +otential difference of '' V

") none of the abo*e

#olution: #eries sameI.

Bulb ratings:2V

RP

= : differentP% same V differentR.

#eries circuit: 2P I R= :

differentR% sameI differentPdifferent brightness.

V I R= : differentR% sameI different V.Ans: "

,. The +ositi*e terminals of two batteries with emf-s of 1 and 2% res+ecti*ely% are

connected together. ere 1 / 2.The circuit is com+leted by connecting the negati*e

terminals. f each battery has an internal resistance of r% the rate in watts at which

electrical energy is con*erted to chemical energy in the smaller battery is:

A)2

1 r


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B)2

1 2r

C) ( )2 1 1 r

!) ( )2 1 1 2r

")2

2 2r

#olution: All elements are in series:

2 1

2I

r

= into battery 1 ).

2 1

1 12

P Ir

= = elect. to chem.)

Ans: !

. The resistance of resistor 1 is twice the resistance of resistor 2. The two are

connected in series and a +otential difference is maintained across the combination.

Then:

A) the current in 1 is twice that in 2

B) the current in 1 is half that in 2

C) the +otential difference across 1 is twice that across 2

!) the +otential difference across 1 is half that across 2

") none of the abo*e are true

#olution: 1 1 2 22 2V IR IR V = = =

Ans: C

'. A battery of emf 2 V is connected to a 3$resistor. As a result% current of , A

e4ists in the resistor. The terminal +otential difference of the battery is:

A) 0

B) 3 V

C) 12 V

!) 15 V

") 2 V

#olution: , 3 15V V= =


3/7

Ans: !

3. The current in the '.0$resistor in the circuit shown is:

A) 0.2 A

B) 0.36 A

C) 1.' A

!) 2. A

") ,.0 A

#olution: 7++er branch:3 12

53 12

UR

= + =

+

8ower branch: , ' 5LR = + =

Total:1

5 2

R= =

12,

I A= =

'1.'

2

II A

= =

Ans: C

6. n the diagrams% all light bulbs are identical and all emf de*ices are identical. n

which circuit % % % V% V) will the bulbs be dimmest9

A)


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B)

C)

!) V

") V

#olution:2V

PR

= :

( )2

2,1

2 2 I

P pR R

= = =

( )2

21

, , II

P pR

= =

( )2

1

2 2III

P pR

= =

( )2

1 1

2 2 IV

P pR

= =

( )2

21

2 2V

P pR

= =

Ans: !

5. A 120$V +ower line is +rotected by a 1'$A fuse. (hat is the ma4imum number of

&120 V% '00 (& light bulbs that can be o+erated at full brightness from this line9

A) 1

B) 2

C) ,

!)

") '

#olution: ma4 120 1' 1500P W= =

ma4 ,1'0

Pn Floor

= =

Ans: C

. n the figure% *oltmeter V1reads 300 V% *oltmeter V2reads '50 V% and ammeter A

reads 100 A. The +ower wasted in the transmission line connecting the +ower houseto the consumer is:

A) 1 ;(

B) 2 ;(

C) '5 ;(


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!) ' ;(

") 30 ;(

#olution: ( ) ( )1 2 100 300 '50 2P I V V kW= = =

Ans: B

10. A certain *oltmeter has an internal resistance of 10%000 and a range from 0 to

12 V. To e4tend its range to 120 V% use a series resistance of:

A) 1%111

B) 0%000

C) 100%000

!) 105%000

") 120%000

#olution: internal resistance is in +arallel to ideal *oltmeter:

ma4

12 120

10%000 10%000I

R= =

+ 10%000 100%000R+ =

0%000R=

Ans: B

11.


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!) % 2% 1% ,

") ,% 1% 2%

#olution: #ince ca+acitor is initially uncharged% 0V= across it when # is closed.

I R=

13I A= % 2 ,I A= % , 12I A= % 2I A=

>an;% least to greatest: %2%1%,

Ans: !

12. n the circuit shown% the ca+acitor is initially uncharged. At time t= 0% switch #

is closed. f denotes the time constant% the a++ro4imate current through the ,

resistor when t= 10 is:

A) 0.,5 A

B) 0.'0 A

C) 0.6' A

!) 1.0 A

") 1.' A

#olution: 110

0 0

tI I e I e = =

0

10 10

3 , I = =

+0.110 1.0

I e A=

Ans: !

1,. A charged ca+acitor is being discharged through a resistor. At the end of one time

constant the charge has been reduced by 1 @ 1e) = 3, of its initial *alue. At the end

of two time constants the charge has been reduced by what +ercent of its initial *alue9

A) 52

B) 53

C) 100

!) between 0 and 100

") need to ;now more data to answer the uestion


7/7

#olution: 1 @ 1e) = 0.3, 1 0.,6e=

( )22

0

1 1 0.,6 53I

eI

= = =

Ans: B

1. A certain ca+acitor% in series with a resistor% is being charged. At the end of 10 ms

its charge is half the final *alue. The time constant for the +rocess is about:

A) 0., ms

B) 2., ms

C) 3. ms

!) 10 ms

") 1 ms

#olution: 0.01

0

1

2

Ie

I

= = 0.01

0.01ln 2

s= =

Ans: "

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