Showing posts with label Electric Current. Show all posts
Showing posts with label Electric Current. Show all posts
Saturday, 14 January 2012
Thursday, 12 January 2012
Direct Current Electrical Motor(Animation)
This Java applet shows a direct current electrical motor which is reduced to the most important parts for clarity. Instead of an armature with many windings and iron nucleus there is only a single rectangular conductor loop; the axis the loop rotates on is omitted.
The red arrows indicate the conventional direction of current (from plus to minus). You can recognize the magnetic field lines (directed from the red painted north pole to the green painted south pole) by the blue color. The black arrows represent the Lorentz force which is exerted on a current-carrying conductor in the magnetic field.
The mentioned Lorentz force is orthogonal to the direction of current and to the magnetic field lines. The orientation of this force results from the well-known three finger rule (for the right hand!)
For Animation CLICK HERE
The red arrows indicate the conventional direction of current (from plus to minus). You can recognize the magnetic field lines (directed from the red painted north pole to the green painted south pole) by the blue color. The black arrows represent the Lorentz force which is exerted on a current-carrying conductor in the magnetic field.
The mentioned Lorentz force is orthogonal to the direction of current and to the magnetic field lines. The orientation of this force results from the well-known three finger rule (for the right hand!)
For Animation CLICK HERE
Potentiometer(Animation)
In this animation you can change position of the sliding contact.You can observe the changes in output voltages and current.
For Animation CLICK HERE
For Animation CLICK HERE
Sunday, 1 January 2012
WHEATSTONE BRIDGE
WHEATSTONE BRIDGE
Wheatstone bridge is an arrangement of four resistances which can be used to measure one of them in terms of rest. Here arms AB and BC are called ratio arm and arms AC and BD are called conjugate arms.
Balanced Wheatstone bridge: The bridge is said to be balanced when deflection in galvanometer is zero i.e. no current flows through the galvanometer or in other words VB = VD. In the balanced condition P/Q = R/S, on mutually changing the position of cell and galvanometer this condition will not change.
Unbalanced Wheatstone bridge: If the bridge is not balanced current will flow from D to B if VD > VB i.e. (VA – VD) < (VA – VB) which gives PS > RQ.
Applications of Wheatstone bridge: Meter bridge, post office box and Carey Foster bridge are instruments based on the principle of Wheatstone bridge and are used to measure unknown resistance.
METER BRIDGE: In case of Meter Bridge, the resistance wire AC is 100 cm long. Varying the position of tapping point B, bridge is balanced. If in balanced position of bridge AB = l,
BC = (100 – l)
So that .Q/P = (100–l)/l
Also P/Q = R/S ⇒ S = (100–l)/l R
Solved example 1: In Wheatstone bridge P = 9 ohm, Q = 11 ohm, R = 4 ohm and S = 6 ohm. How much resistance must be put in parallel to the resistance S to balance the bridge
(A) 24 ohm (B) 44/9 ohm (C) 26.4 ohm (D) 18.7 ohm
Solution: (C) (For balancing bridge)
⇒ S' = 4×11/9 = 44/9 ⇒ 1/S' = 1/r + 1/6
⇒ 9/44 – 1/6 = 1/r ⇒ r = 132/5 = 26.4 Ω
Solved example 2: A voltmeter having a resistance of 998 ohms is connected to a cell of emf 2 volt and internal resistance 2 ohm. The error in the measurement of emf will be
(A) 4 ×10–1 volt (B) 2 ×10–3 volt
(C) 4 ×10–3 volt (D) 2 ×10–1 volt
Solution: (C) Error in measurement = Actual value – Measured value
Actual value = 2A
i = 2/998+2 = 1/500 A
Since E = V + ir = ⇒ V = E – ir = 2 – 1/500 × 2 = 998/500 V
Measured value = 998/500 V ⇒ Error = 2 – 998/500 = 4 × 10–3 volt.
Potentiometer
Potentiometer is a device mainly used to measure emf of a given cell and to compare emf's of cells. It is also used to measure internal resistance of a given cell.
Circuit diagram: Potentiometer consists of a long resistive wire AB of length L (about 6 m to 10 m long) made up of mangnine or constantan and a battery of known voltage e and internal resistance r called supplier battery or driver cell. Connection of these two forms primary circuit.
One terminal of another cell (whose emf E is to be measured) is connected at one end of the main circuit and the other terminal at any point on the resistive wire through a galvanometer G. This forms the secondary circuit. Other details are as follows
J = Jockey
K = Key
R = Resistance of potentiometer wire,
r = Specific resistance of potentiometer wire.
Rh = Variable resistance which controls the current through the wire AB
(i) The specific resistance (r) of potentiometer wire must be high but its temperature coefficient of resistance (a) must be low.
(ii) All higher potential points (terminals) of primary and secondary circuits must be connected together at point A and all lower potential points must be connected to point B or jockey.
(iii) The value of known potential difference must be greater than the value of unknown potential difference to be measured.
(iv) The potential gradient must remain constant. For this the current in the primary circuit must remain constant and the jockey must not be slided in contact with the wire.
(v) The diameter of potentiometer wire must be uniform everywhere.
Potential gradient (x): Potential difference (or fall in potential) per unit length of wire is called potential gradient i.e. x = V/L volt/m where V = iR = (e/R+Rn+r)R.
So x = V/L = iR/L = ip/A = e/(R+Rh+r) . R/L
(i) Potential gradient directly depends upon
(a) The resistance per unit length (R/L) of potentiometer wire.
(b) The radius of potentiometer wire (i.e. Area of cross-section)
(c) The specific resistance of the material of potentiometer wire (i.e. r)
(d) The current flowing through potentiometer wire (i)
(ii) Potential gradient indirectly depends upon
(a) The emf of battery in the primary circuit (i.e. e)
(b) The resistance of rheostat in the primary circuit (i.e. Rh)
Working: Suppose jockey is made to touch a point J on wire then potential difference between A and J will be V = xl
At this length (l) two potential difference are obtained
(i) V due to battery e and
(ii) E due to unknown cell
If V > E then current will flow in galvanometer circuit in one direction 
If V < E then current will flow in galvanometer circuit in opposite direction 
If V = E then no current will flow in galvanometer circuit this condition to known as null deflection position, length l is known as balancing length.
In balanced condition E = xl
or E = xl = V/L l = iR/L l = (e/R+Rh+r) × R/L × l
If V is constant then L ∝ l ⇒ x1/x2 = L1L2 = l1/l2
Standardization of Potentiometer: The process of determining potential gradient experimentally is known as standardization of potentiometer.
Let the balancing length for the standard emf E0 is l0 then by the principle of potentiometer E0 = xl0 ⇒ x = E0/l0
Sensitivity of potentiometer: A potentiometer is said to be more sensitive, if it measures a small potential difference more accurately.
(i) The sensitivity of potentiometer is assessed by its potential gradient. The sensitivity is inversely proportional to the potential gradient.
(ii) In order to increase the sensitivity of potentiometer
(a) The resistance in primary circuit will have to be decreased.
(b) The length of potentiometer wire will have to be increased so that the length may be measured more accuracy.
Difference between voltmeter and potentiometer
Voltmeter | Potentiometer |
It's resistance is high but finite | It's resistance is infinite |
It draws some current from source of emf | It does not draw any current from the source of unknown emf |
The potential difference measured by it is lesser than the actual potential difference | The potential difference measured by it is equal to actual potential difference |
Its sensitivity is low | Its sensitivity is high |
It is a versatile instrument | It measures only emf or potential difference |
It is based on deflection method | It is based on zero deflection method |
Bulbs in series and parallel for CET aspirants
Combination of Bulbs
Bulbs in Series
(i) Total power consumed 1/Ptotal = 1/P1 + 1/P2 +...
(ii) If ‘n’ bulbs are identical, Ptotal = P/N
(iii) Pconsumed (Brightness) ∝ V ∝ R ∝ 1/Prated i.e. in series combination bulb of lesser wattage will give more bright light and p.d. appeared across it will be more.
Bulbs in Parallel
(i) Total power consumed
Ptotal = P1 + P2 + P3 +...+ Pn
(ii) If ‘n’ identical bulbs are in parallel Ptotal = nP
(iii) Pconsumed (Brightness) ∝ PR ∝ i ∝ 1/R i.e. in parallel combination, bulb of greater wattage will give more bright light and more current will pass through it.
Solved example 1: An electric bulb is rated 220 volt and 100 watt. Power consumed by it when operated on 110 volt is
(A) 50 watt (B) 75 watt (C) 90 watt (D) 25 watt
Solution: (D) Resistance of the bulb V2/PRotate = 220×220/100 = 484 Ω
When connected with 110 V, the power consumed
Pconsumed = V2/R = 110×110/484 = 25W
Solved example 2: Two bulbs are working in parallel order. Bulb A is brighter than bulb B. If RAB are their resistance respectively then and R
(A) RA > RB (B) RA < RB
(C) RA = RB (D) None of these
Solution: (B) In parallel Pconsumed ∝ Brightness ∝ 1/R
PA > PB (given). RA < RB
Heating Effect of Electric Current
Heating Effect of Current
Joule Heat
When some potential difference V is applied across a resistance R then the work done by the electric field on charge q to flow through the circuit in time t will be 
Joule. This work appears as thermal energy in the resistor.
Joule. This work appears as thermal energy in the resistor.Heat produced by the resistance R is 
Cal. This relation is called joules heat.
Cal. This relation is called joules heat.Electric Power
The rate at which electrical energy is dissipated into other forms of energy is called electric power i.e.
Units: It’s S.I. unit is Joule/sec or Watt
Bigger S.I. units are KW, MW and HP, remember 1 HP = 746 Watt
Rating values
On electrical appliances (Bulbs, Heater, Geyser … etc). Wattage, voltage, … etc. are printed called rating values e.g. If suppose we have a bulb of 40 W, 220 V then rated power (PR) = 40 W while rated voltage (VR) = 220 V.
Resistance of electrical appliance
If variation of resistance with temperature is neglected then resistance of any electrical appliance can be calculated by rated power and rated voltage i.e. by using 
Power consumed (illumination)
An electrical appliance (Bulb, heater, … etc.) consume rated power (PR) only if applied voltage (VA) is equal to rated voltage (VR) i.e. If VA = VR
So Pconsumed = PR. If VA < VR then Pconsumed = VzA/R also we have R = VzR/PR
So Pconsumed (Brightness) = (V2a/V2R) × PR
Long distance power transmission
When power is transmitted through a power line of resistance R, power-loss will be i2R
Now if the power P is transmitted at voltage V then P = Vi , i.e. i = (P / V)
So Power loss = P2/V2 × R
Now as for a given power and line, P and R are constant so Power loss ∝ (1/V2)
So if power is transmitted at high voltage, power loss will be small and vice-versa. This is why long distance power transmission is carried out at high voltage.
Electricity Consumption
1. The price of electricity consumed is calculated on the basis of electrical energy and not on the basis of electrical power.
2. The unit Joule for energy is very small hence a big practical unit is considered known as kilowatt hour (KWH) or board of trade unit (B.T.U.) or simple unit.
3. 1 KWH or 1 units is the quantity of electrical energy which dissipates in one hour in an electrical circuit when the electrical power in the circuit is 1 KW thus 1KWH = 1000W × 3600 sec = 3.6 × 106 J.
4. Important formulae to calculate the no. of consumed units is
n = Total Watt × Total Hours/1000
Solved example 1: Two heater wires of equal length are first connected in series and then in parallel. The ratio of heat produced in the two cases is
(A) 2 : 1 (B) 1 : 2 (C) 4 : 1 (D) 1 : 4
Solution: (D) Power consumed means heat produced.
For constant potential difference Pconsumed = Heat ∝ 1/Req
H1/Hz = Rz/R1 = R/2/2R = 1/4 (Since Rz = R.R./R+R = R/2 and R1 = R + R = 2R)
Solved example 2: A wire when connected to 220 V mains supply has power dissipation P1. Now the wire is cut into two equal pieces which are connected in parallel to the same supply. Power dissipation in this case is P2. Then P2 : P1 is
(A) 1 (B) 4 (C) 2 (D) 3
Solution: (B) When wire is cut into two equal parts then power dissipated by each part is 2P1
So their parallel combination will dissipate power
P2 = 2P1 + 2P1 = 4P1, Which gives P2/P1 = 4.
Wednesday, 28 December 2011
The animation will discuss the relation between resistance length,resistivity and area.In that animation You can change the length area and resistivity value using movable switch given there .Try to play with that.And get a feel on how exactly it changes.
Animation of Variation of resistance
Animation of Variation of resistance
