PHYSICS 120:ELECTRICITY AND MAGNETISM

TUTORIAL QUESTIONS

DC CIRCUITS

Question 61

A 1000 Ω 2.0 Wresistor is needed but only 1000 Ω 1.0 Wresistors are available.How can the

required resistance and power rating be obtained by a combinationof the available units?

What power is then dissipated in each resistor?

The current through the resistor can be found by

P = V I = I

2

R =

V

2

R

so the current through the 2.0 Wresistor would be

I =

P

R

=

2.0

1000

=

1

10

√

5

A

The maximum current that can ﬂow through the 1.0 Wresistor is

I =

1.0

1000

=

1

10

√

10

A

hence the 1.0 Wresistors must be wired in parallel.For a maximum current of

1

10

√

5

A in the

circuit,two resistors in parallel each would carry

1

2

1

10

√

5

<

1

10

√

10

A,so this is permissible.

Two 1000 Ω resistors in parallel would result in an equivalent resistance of

R

EQ

=

1000 ×1000

1000 +1000

= 500 Ω

so two combinations like this in series results in an equivalent resistance of 1000 Ω.The

power dissipated in each resistor is

P = I

2

R =

1

20

√

5

2

×1000 = 0.50 W

Question 62

Each of the three resistors in the ﬁgure has a resistance of 2.0 Ω and can dissipate a maxi-

mum of 18 W without becoming excessively heated.What maximum power can the circuit

dissipate?

page 1 of 6

PHYS120 ELECTRICITY AND MAGNETISM:TUTORIAL QUESTIONS

The maximum current that can ﬂow in the circuit is

I =

P

R

=

18

2.0

= 3.0 A

since this is the current ﬂowing in the single 2.0 Ω resistor.The equivalent resistance for the

whole circuit is

R

EQ

=

2.0 ×2.0

2.0 +2.0

+2.0 = 3.0 Ω

Hence the maximum power dissipated is

P = I

2

R = 3.0

2

×3.0 = 27 W

Question 63

(a) Find the potential diﬀerence V

ad

in the circuit

(b) Find the potential diﬀerence across the 4.00 V cell.

(c) A battery of emf 17.0 V and internal resistance 1.00 Ω is inserted in the circuit at d

with its positive terminal connected to the positive terminal of the 8.00 V battery.Find

V

bc

between the terminals of the 4.00 V battery.

(a) Choosing a direction for the current as anti-clockwise and a direction for the loop as

dcbad,Kirchhoﬀ’s second rule gives

−9I −0.5I +4 −6I −8I +8 −0.5I = 0

This leads to a current of

I =

12

24

= 0.50 A

To ﬁnd V

ad

,start at point a and move in the direction of the current to point d:

V

a

−8I +8 −0.5I = V

d

The potential diﬀerence V

ad

= V

d

−V

a

is then

V

ad

= V

d

−V

a

= 8 −(8 ×0.5) −(0.5 ×0.5) = 3.75 V

(b) The potential diﬀerence V

cb

is found by starting at point c and moving to point d:

V

c

−0.5I +4 = V

b

∴ V

cb

= 4 −0.25 = 3.75 V

page 2 of 6

PHYS120 ELECTRICITY AND MAGNETISM:TUTORIAL QUESTIONS

(c) With a new battery inserted in the circuit,the Kirchhoﬀ II equation becomes

−9I −0.5I +4 −6I −8I +8 −0.5I −17 −(1 ×I) = 0

This now leads to a current of

I = −

5

25

= −0.20 A

The potential diﬀerence V

cb

is then

V

cb

= 4 −(0.5 ×I) = 4 −(0.5 ×(−0.20)) = 4.10 V

Question 64

A dry cell having an emf of 1.55 V and an internal resistance of 8.00×10

−2

Ω supplies current

to a 2.00 Ω resistor.

(a) Determine the current in the circuit.

(b) Calculate the terminal voltage of the cell.

(a) The current in the circuit is given by

1.55 = (2.00 +0.08) ×I ∴ I =

1.55

2.08

= 0.745 A

(b) The terminal voltage is the voltage drop across the 2.00 Ω resistor:

terminal voltage = 2.00 ×0.745 = 1.49 V

Question 65

How many cells,each having an emf of 1.5 V and an internal resistance of 0.50 Ω must be

connected in series to supply a current of

5

3

A to operate an instrument having a resistance

of 6.0 Ω?

page 3 of 6

PHYS120 ELECTRICITY AND MAGNETISM:TUTORIAL QUESTIONS

Suppose there are n cells,then the current in the circuit,the total emf and the total resistance

is given by

1.5n = (0.5n +6.0) ×

5

3

∴ n =

6

(

3

5

×1.5 −0.5)

=

6

0.9 −0.5

= 15

Question 66

A battery has an internal resistance of 0.50 Ω.A number of identical light bulbs,each with

a resistance of 15 Ω,are connected in parallel across the battery terminals.The terminal

voltage of the battery is observed to be one-half the emf of the battery.How many bulbs

are connected?

Suppose there are n light bulbs.The total resistance for n light bulbs in parallel is

1

R

EQ

=

1

15

+

1

15

+

1

15

+...=

n

15

∴ R

EQ

=

15

n

The terminal voltage V is

V = IR

EQ

=

1

2

E

where E = (R

EQ

+0.50)I.Hence

IR

EQ

=

1

2

(R

EQ

+0.50)I

∴

1

2

R

EQ

=

1

2

(0.5)

R

EQ

= 0.5

⇒n =

15

0.5

= 30

page 4 of 6

PHYS120 ELECTRICITY AND MAGNETISM:TUTORIAL QUESTIONS

Question 68

Find the magnitude and direction of the current in the 2.0 Ω resistor in the diagram.

Using Kirchhoﬀ’s rules,the equations needed to ﬁnd the currents are:

KI at a:I

1

+I

2

+I

3

= 0 (I)

KII for loop abcda:−2I

2

−1 −4 +3I

3

= 0 (II)

KII for loop aefba:−I

1

+1 +2I

2

= 0 (III)

(I) gives:I

3

= −I

1

−I

2

(IV)

substitute (IV) into (II):−2I

2

−5 +3(−I

1

−I

2

) = 0

−3I

1

−5I

2

−5 = 0 (V)

multiply (III) by 3:−3I

1

+6I

2

+3 = 0 (VI)

(V) - (VI):−11I

2

−8 = 0

∴ I

2

= −

8

11

A

Hence I

2

= −0.73 A is the current through the 2.0 Ω resistor and it ﬂows from b to a.

Question 69

Determine the voltage across the 5.0 Ω resistor in the circuit below.Which end of the resistor

is at the higher potential?

The Kirchhoﬀ equations are

KI at a:I

1

+I

2

= I

3

(I)

KII for loop abcda:10I

2

−15 +2 +10I

3

= 0 (II)

KII for loop adefa:−10I

3

−2 +10 −5I

1

= 0 (III)

page 5 of 6

PHYS120 ELECTRICITY AND MAGNETISM:TUTORIAL QUESTIONS

These are solved simultaneously as follows:

rearrange (I):I

2

= I

3

−I

1

(IV)

substitute (IV) into (II):10(I

3

−I

1

) −13 +10I

3

= 0

20I

3

−10I

1

−13 = 0 (V)

multiply (III) by 2:−20I

3

−10I

1

+16 = 0 (VI)

(V) + (VI):−20I

1

+3 = 0

∴ I

1

=

3

20

A (VII)

The voltage drop across the 5.0 Ω resistor is

V = I

1

R =

3

20

×5.0 =

3

4

= 0.75 V

Current I

1

is positive,therefore it ﬂows in the direction of f to a.That means that point a

is at a lower potential than point f.

page 6 of 6

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