### RESISTANCES IN SERIES AND PARALLEL

RESISTANCES IN SERIES AND PARALLEL

1. Resistances in Series:-

In a circuit, resistors connected end-to-end are said to be in series, if the same current exists in all of them through a single path.

Note:-

When resistors are connected in series

1. The same current passes through all of them 2. The voltage of the Battery is divided among them. 3. The equivalent resistance of the combination is equal to the sum of individual resistances 4. The equivalent resistance of the combination is more than that of any of the individual resistances.

2. Resistances in parallel:-

In a circuit, resistors connected to common terminals are said to be in parallel, if identical potential difference exists across all of them.

Note:-

When resistors are connected in parallel.
1. The total current of the circuit is divided among them. 2. Same potential difference exist across all of them 3. The reciprocal of the equivalent resistance is equal to sum of reciprocals of the individual resistance 4. The equivalent resistance of parallel combination is less than that of any individual resistance.

3. Show that effective resistance of a series combination in a circuit is equal to the sum of their resistances.
Or
Derive R=R1+R2 + R3

In series combination total potential difference is equal to sum of individual potential difference. 4. Show that the reciprocal of the effective resistance of a parallel combination in a circuit is equal to the sum of the reciprocals of individual resistances.
Or  II. Problems:-

1. What is the equivalent resistance of two resistors
when connected in a) Series.
b) Parallel. 2. Calculate the equivalent resistance of two resistors of 100Ω and 1Ω connected in parallel.

Given, 3. Two conductors of resistances 20 Ohms and 40 Ohms are connected in parallel to a 4V battery. Draw the circuit diagram. Find the Current in the circuit and also in each of the conductors.

Given, i=0.2+0.1
i=0.3 Amperes.
Total current in the circuit =0.3 Amperes.

4. The effective resistance of combination of 3 resistors in series is 100 Ohms. The value of 2 resistors are 20 Ohms and 30 Ohms. Find the value of 3rd resistance?  5. Two coils have an effective resistance of 9Ω when connected in series and 2 Ω when connected in parallel. Find the values of individual resistance of coil

Let R1 and R2 be the resistances of two coils.
a) When they are connected in series HEATING EFFECTS OF ELECTRIC CURRENT : JOULE’S LAW

1. Joule Heat:-

The heat developed due to conversion of any form of work or energy is called Joule heat.

Note:-

The quantity of heat produced by a current in a given time is directly proportional to resistance.
QαR

The quantity of heat produced by a current in a resistance is directly proportional to the time for which current flows through it.
Qαt

The quantity of heat produced by a current in a resistance is directly proportional to the square of the magnitude of the current.
Qαi2

2. Joule’s Law:-

Joule’s law states that a given amount of work done in different ways produces the same quantity of heat in all cases.
Or

A given amount of work done in different ways produces the same quantity of heat in all cases and is directly proportional to heat produced.
WαQ The potential difference between two points in an electric circuit is the work done in moving unit positive charge from one point to other. 4. Show that electrical energy W=i2Rt. 5. Describe an experiment to determine Joule’s Constant(J). Parts:-

B-battery, V-Voltmeter, K-plug key, Rh-Rheostat, A-Ammeter, H-coil,
Th-Thermometer, C-copper calorimeter.

1. Take a copper calorimeter (C) with stirrer (S) enclosed in an insulating wooden box (I).

Determine the mass of calorimeter with stirrer. Let its weight be W1 gm.

Pour water into calorimeter up to half of its volume and weigh it. Let the weight be W2 gm.

The mass of water (m) is equal to (W2-W1) gm.

Record the initial temperature of water t10c with the help of a thermometer.

Immerse a heating coil (H) in that calorimeter.

Connect the two ends of heating coil in a circuit containing a battery B, Plug-key(K), Ammeter (A) and a Rheostat(Rh) in series as shown in the figure.

Connect a Voltmeter (V) parallel to the coil to determine the potential difference (V) across it.

Pass the electric current ‘i’ through the coil for time ‘t’ seconds.
Stir the water such that the temperature of calorimeter and its contents rises at least by 50c.

Observation:- 6. What do you understand by the term ‘wattage’ of an electrical appliance?

Wattage of an electrical appliance is defined as the rate at which electrical energy is consumed by it.

Wattage of a bulb or an immersion heater is the power consumed by it and is expressed in watts.

7. Define the mechanical equivalent of a heat or Joule’s constant

The work required to be done to produce a quantity of heat of 1 calorie is 4.18 Joules. This is called mechanical equivalent of heat or Joule’s constant.

8. Define watt-hour. What is a kilowatt Hour (KWH)

Watt-hour is defined as the electrical energy consumed by an appliance of 1 watt for an hour.
The units of house hold consumption of electrical energy is Kilo-watt hour.

9. Why is heat produced when current passes through a Resistance

When a potential difference is applied between the two ends of metallic conductor, the free electron gain energy and begin to drift. In this process they collide with the ion core of the conductor and transfer energy. Then the ions begin to vibrate with increased amplitude. This results in the increase of temperature of the conductor which transforms the heat to the surrounding.

Note:-

1. Let there be four tube lights each of 40 watts in the house. These are used on average for 5 hours per day. In addition, let there be an immersion heater of 1500 watts used on an average for one hour per day. Then, how many units of electricity are consumed in a month?

Energy consumed by the tube light in a day= Number of tube lights x wattage x hours of use per day.
=4x40x5
=800 WH.
Energy consumed by the immersion heater in a day =1x1500x1=1500WH.
Total energy consumed in a day=800+1500=2300WH. 2. An electric heater of resistance 23 Ohms is used on a mains supply of 230 volts. Calculate the amount of heat given out by it in a minute (J=4.2 Joules/cal)

Given,
V= 230 V
R=23Ω
t=1 minute=60 seconds. 3. A house is fitted with 10 lamps of each 60 watts. If each lamp burns for 5 hours a day on an average, find the cost of consumption in a month of 30 days at 80 paise per unit.

Electric energy consumed by 10 lamps at the rate of 5 hours per day for 30 days in KWH 5. A potential difference of 230 v is applied between the terminals of an electric lamp. A current of 0.5 Amperes passes through it.
a) Calculate its resistance and wattage.
b) If it is used for 100 hours how many units of energy are consumed?

Given,
Potential difference=V=230 Volts.
Current=i=0.5 Amperes. Resistance=R=460Ω

Wattage =v x i (voltage x Current)
=230 x 0.5
Wattage=115 watts.

b) Energy consumed =11.5 units
or
=11.5 KWH
Energy consumed=11.5 KWH or 11.5 units.

6. An electric installation consists of 100 lamps, each drawing 0.2 A at 220 V supply. Find the cost of working of the installation for a month of 30 days of 5 hours a day, if energy is charged at the rate of 2 rupees per unit.

Given,
Current=i=0.2 Amperes.
Voltage=V=220 volts.
Number of lamps=100
Time of use=5 hours.
Wattage= voltage x current
=220 x 0.2
=44 watts.
Energy consumed by 100 lamps at the rate of 5 hours per day for 30 days in KWH =44 x 15
=660 KWH or 660 units.
For one unit cost of working=2 rupees.
For 660 units cost of working=660 x 2
=1320 rupees.
Cost of working of installation=1320 Rupees.