VII. DC Circuits - Worked Examples

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics

8.02

Spring 2003

VII. DC Circuits - Worked Examples

Example 1: Equivalent resistance

Consider

the

circuit

shown

below.

For

a

given

resistance

R 0

,

what

must

be

the

value

of

R 1

so that the input resistance between the terminals is equal to R0 ?

Solution:

The equivalent resistance, R', due to the three resistors on the right is

1 R'

=

1 R

1

+

R 0

1 +

R 1

=

R0 + 2R1

R 1

(

R 0

+

R 1

)

(1.1)

or

R'

=

R1 ( R0 + R1 )

R 0

+

2R

(1.2)

Since R' is in series with the fourth resistor R1, the equivalent resistance of the entire configuration becomes

If

Req

=

R 0

,

then

( ) Req

=

R 1

+

R1 R0

R 0

+

+ R1

2

R 1

=

3R12 R

0

+ 2R1R0 + 2R1

(1.3)

( ) R 0

R 0

+

2

R 1

=

3R12

+

2

RR 10

R2 0

=

3R12

(1.4)

or

R 1

=

R 0 3

(1.5)

1

Example 2: Variable resistance

Show that, if a battery of fixed emf and internal resistance Ri is connected to a variable external resistance R , the maximum power is delivered to the external resistor when R = Ri

Solution:

Using Kirchhoff's rule,

= I ( R + Ri )

(1.1)

which implies

I

=

R

+

Ri

(1.2)

The power dissipated is equal to

P = I2R = 2 R

( R + Ri )2

(1.3)

To find the value of R which gives out the maximum power, we differentiate P with respect to R and set it to be 0:

dP dR

=

2

(

R

1 + Ri

)2

-

(R

2R + Ri

)2

=2

Ri - R

( R + Ri )3

=

0

(1.4)

which implies

R = Ri

(1.5)

2

Example 3: Multiloop Circuit Find the currents I1, I2 and I3 in the circuit below.

Solution:

Since there are three unknowns in the system, we need three linearly independent equations to find the solutions.

Applying Kirchhoff's current rule to junction c yields

I 1

+

I 2

=

I 3

(2.1)

The other two equations can be obtained by using the voltage rule, which states that the

net potential difference across all elements in a closed circuit loop is zero. Traversing the loops befcb and abcda clockwise leads to

- 2

-

IR 22

- 1

+

IR 11

=

0

(2.2)

and

2

-

IR 11

-

IR 33

=

0

One may also consider loop abefcda, which gives

(2.3)

-

I

2

R 2

- 1

-

IR 33

=

0

(2.4)

However, the above equation is not linearly independent since it is simply the sum of the previous two.

The solutions to the above three equations are given by

3

I 1

=

1R3 RR

12

+ 2R3

+

RR 13

+ 2R2

+

RR 23

I 2

=

-

1R1 RR

12

+

1

R 3

+

2

R 3

+

RR 13

+

RR 23

(2.5)

I 3

=

2

R 2

-

1R1

RR 12

+

RR 13

+

RR 23

From the above expression, we see that I2 is a negative quantity. This simply implies that the direction of I2 is opposite of what we have initially assumed.

Example 4: RC circuit

In the circuit below, suppose the switch has been open for a very long time. At t = 0, it is suddenly closed.

(a) What is the time constant before the switch is closed?

(b) What is the time constant after the switch is closed?

(c) Find the current through S as a function of time after the switch is closed.

Solution:

(a) Before the switch is closed, the two resistors R1 and R2 are in series with the capacitor.

Since

the

equivalent resistance

is

Req

=

R 1

+

R 2

,

the

time

constant is

given

by

=

ReqC

=

(

R 1

+

R 2

)C

(3.1)

The amount of charge stored in the capacitor is

4

q(t) = C (1- e-t / )

(3.2)

(b) After the switch is closed, the closed loop on the right becomes a decaying RC circuit with time constant ' = R2C . Charge begins to decay according to

q '(t) = C e-t / '

(3.3)

(c) The current passing through S consists of two sources: the steady current I1 from the left circuit, and the decaying current I2 from the RC circuit. The currents are given by

I 1

=

R

1

I

'(t)

=

dq ' dt

=

-

C '

e-t / '

=

-

R

2

e-

t

/

R2C

(3.4)

The negative sign in I'(t) indicates that the direction of flow is opposite of the charging process. Thus, since both I1 and I' move downward across the switch S, the total current

is

I (t)

=

I1

+

I

'(t)

=

R1

+

R2

e-

t

/

R2C

(3.5)

Example 5: Parallel vs. series connections

The figures below show two resistors with resistance R1 and R2 connected in parallel and in series. The battery has a terminal voltage of .

Suppose R1 and R2 are connected in parallel. (a) Find the power delivered to each resistor.

5

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