푣표푙푡푎푔푒

resistive effect offered by a collection of

resistive device which are present in a

circuit.

푐푢푟푟푒푛푡 =

푟푒푠푖푠푡푎푛푐푒

2푉

푐푢푟푟푒푛푡 =

8Ω

Example 07

푐푢푟푟푒푛푡 = 0.25퐴

Several 150Ω resistors are to be connected

so that a current of 2A flows from a 50V

source. How many resistors are required and

how should they be connected?

(ii) When switches 퐾₁and 퐾₂are both

closed

Solution

Step 1: To find the total resistance

푣표푙푡푎푔푒

푅_퐸=

푐푢푟푟푒푛푡

50푉

푅_퐸=

2퐴

Note When switches 퐾₁and 퐾₂are

푅_퐸= 25Ω

The resistors must be connected in

both closed

the 5Ω resistor

is bypassed since the wire has low

resistance hence will allow current

to pass through it.

parallel since the 푅_퐸must be, lower than

150Ω

Resistance = 3Ω

Step 2: To find the total number of

resistors

푣표푙푡푎푔푒

푐푢푟푟푒푛푡 =

By using the equation for similar parallel

resistors

푟푒푠푖푠푡푎푛푐푒

1

2푉

푐푢푟푟푒푛푡 =

3Ω

= 푛 (

)

25

150

Where n is the number of resistors

푐푢푟푟푒푛푡 = 0.67퐴

(iii)When switch 퐾₁is open and 퐾₂is

closed the circuit is said to be open.

In an open circuit NO current flows.

∴ 푐푢푟푟푒푛푡 = 0

150

푛 =

25

푛 = 6

Example 08

Calculate the current across each resistor in

the circuit diagram below.

Example 06

What is effective resistance?

Answers

Effective resistance: -is the amount of

resistance which a single device will need in

order to replace (equal to) the overall

Explain why resistance cannot be accurately

measured using just a voltmeter and

ammeter.

Answers

This is because the voltage drop across the

ammeter and the shunting effect of the

voltmeter

Solution

Step 1: To find the total p.d across the

circuit

Example 10

푣표푙푡푎푔푒 = 푐푢푟푟푒푛푡 × 푟푒푠푖푠푡푎푛푐푒

Why it is preferable to connect bulbs in

parallel than in series

8 × 6

푣표푙푡푎푔푒 = 퐼 × (

)

8 + 6

Answers

8 × 6

8 + 6

Connecting bulbs in parallel is advisable for

several practical reasons as follows:

푣표푙푡푎푔푒 = 2퐴 × (

)

푣표푙푡푎푔푒 = 6.86푉

(v)

Bulbs get full voltage

Step 2: To find the current in the 8Ω

In parallel connections, all bulbs

receive the same voltage as the

power source. This ensures each

bulb glows at its full brightness.

resistor

푣표푙푡푎푔푒

푐푢푟푟푒푛푡 =

푟푒푠푖푠푡푎푛푐푒

6.86푉

푐푢푟푟푒푛푡 =

8Ω

(vi)

Independent operations

푐푢푟푟푒푛푡 = 0.857퐴

If one bulb burns out or is

removed,

the

others

keep

working. This is because each

bulb has its own separate circuit

path.

Step 3: To find the current in the 6Ω

resistor

푣표푙푡푎푔푒

푐푢푟푟푒푛푡 =

푟푒푠푖푠푡푎푛푐푒

6.86푉

푐푢푟푟푒푛푡 =

6Ω

(vii)

Consistency brightness

Adding more bulbs in parallel

does not affect the brightness of

푐푢푟푟푒푛푡 = 01.143퐴

others,

unlike

the

series

Example 09

connection where the voltage

gets divided.

(viii) Easier to match the devices

Bulb D will not light up, this is because the

circuit through Bulb C and D is open by the

blown off bulb C.

Different types of bulbs or electrical devices

can be connected in parallel even if they

require different currents, as long as the

voltage is the same.

Concept

When the switch is closed the current flows

up to the junction at which it divides into

two parts, one current flow through A and B

and the other through Bulb C and D. Since

Bulb C is blown off, no current flows in

Bulb C and D. Therefore, Bulbs A and B

maintain the same brightness since the same

current flows in each bulb as they are

connected in series.

Example 11

Why is an ammeter constructed with a low

resistance?

Answers

To allow all the current to be accurately

measured without being affected by the

resistance.

The low resistance ensures that all the

current to be measured passes through the

ammeter without being opposed/resisted.

Example 13

(a) The figure below shows an electric

circuit. Carefully study it and calculate the

value of the current x and r

Example 12

In the figure below, the labels A, B, C and D

are identical electric bulbs connected in a

circuit. Explain what will happen to the

bulbs A, B and D when bulb C blows off

and the key K is closed.

Solution

From the current law “ at any junction

of the circuit, the main current

distributes into two or more individual

currents depending on the number of of

electric

components

connected

in

parallel”

Answers

When key K is closed and C blows off, Bulb

A and B will light up with the same

brightness because the path taken by current

through Bulb A and B is not blocked.

Step 1: To find the value of x

From

Total current = current in x + current in

the 4Ω

Current in x = total current – current in

the 4Ω

1

x = 5.2A − 3.2A

x = 2.0A

=

+

푅

4

6.4

1

6.4 + 4

4 × 6.4

Step 2: To find the value of r

푅

25.6

푅₁and 푅₂have the same voltage since

푅 =

10.4

푅 = 2.46Ω

they are in parallel.

Example 14

푙푒푡 푟 Ω = 푅₁, 400Ω = 푅₂

In the figure below, lamp 퐿₃shines brighter

but lamp 퐿₄is dimmer when switch 푺_ퟐis

closed. Give reason for this.

From

V = I R

I = 3.2A, R = 4.0Ω

(

)

푉 = 3.2 × 4.0 푉

푉 = 12.8푉

Answers

The voltage across the resistor is 12.8V

Because all the current in the circuit

flows through 퐿₃and hence it is

brighter

푉

푟 =

퐼

12.8푉

푟 =

Lamps 퐿₃and 퐿₅are in parallel.

They are identical and hence have

the same resistance. The current

through each of these lamps will be

half that of 퐿₃hence 퐿₄and 퐿₅have

the same brightness but dimmer

than 퐿₃.

2.0퐴

푟 = 6.4Ω

(b) . From part (a) determine the effective

resistance.

푇표푡푎푙 푝. 푑

푅 =

푇표푡푎푙 푐푢푟푟푒푛푡

12.8푉

푅 =

Example 15

5.2퐴

Study carefully the diagram below and

answer the next questions

푅 = 2.46Ω

Alternative solution

Since the resistors are in parallel, the

effective resistance is given by

1

=

+

푅

푅₁

푅₂