Current Electricity Concept Page - 12

Definition

Null point

The null point or balanced point is an arrangement of resistors across the arms of a Wheatstone bridge or meter bridge such that the deflection in the galvanometer is zero. A,B,C and D are four coils of wires of 2,2,2$2,2,2$ and 3$3$ ohm resistances respectively and are arranged to form a Wheatstone bridge.The resistance which the coil 'D' must be shunted in order that the bridge may be balanced is given by:Wheatstone condition:

RARC=RBRD${\displaystyle \frac{R_A}{R_C}}={\displaystyle \frac{R_B}{R_D}}$
RB=RD$R_B=R_D$
RB=3×RS3+RS$R_B={\displaystyle \frac{3\times R_S}{3+R_S}}$
6+2RS=3RS$6+2R_S={3R}_S$
RS=6Ω$R_S=6\mathrm{\Omega }$

Diagram

Describe the process of determination of unknown resistance using meter bridge

Procedure:
The metre bridge, also known as the slide wire bridge consists of a one metre long wire of uniform cross sectional area, fixed on a wooden block. A scale is attached to the block. Two gaps are formed on it by using thick metal strips in order to make the Wheat stones bridge. The terminal B between the gaps is used to connect galvanometer and jockey.
A resistance wire is introduced in gap S and the resistance box is in gap R. One end of the galvanometer is connected to terminal D and its other end is connected to a jockey. As the jockey slides over the wire AC, it shows zero deflection at the balancing point (null point).
If the length AB is l$l$, then the length BC is  ( 100−l$100-l$ ).
Then, according to Wheatstones principle; 
                                                                 XR=l(100−l)${\displaystyle \frac{X}{R}}={\displaystyle \frac{l}{(100-l)}}$ 
Now, the unknown resistance can be calculated as,
                                                                X=Rl(100−l)$X=R{\displaystyle \frac{l}{(100-l)}}$
The specific resistance or resistivity of the material of the wire can be then calculated by using the relation,
                       ρ=πr2XL$\rho ={\displaystyle \frac{\pi r^{2}X}{L}}$ ; Where L$L$ be the length of the wire and r$r$ be its radius.

Definition

Errors and precautions while using meter bridge

Possible errors: Non-uniformity in the wire brings about different resistance per unit length, thus inducing error in unknown resistance. Moreover there may be errors due to the heating effect, end corrections introduced due to shift of the zero of the scale.Precautions:

• The battery key should always be pressed before the galvanometer contact is made on the bridge wire to ensure that a steady current is flowing in the circuit .Otherwise a counter e.m.f. would be self induced somewhere in the circuit which would make it impossible to find the exact balance point.  
• The experiment should be repeated by interchanging the resistances in left and right gaps.
• The value of resistance of the resistance is such that the balance condition is achieved near the center of the wire.

Example

Problems using meter bridge

When an unknown resistance and a resistance of 4Ω$4\mathrm{\Omega }$ are connected in the left and right gaps of a meter bridge, the balance point is obtained at 50cm$50cm$. The shift in the balance point, if a 4Ω$4\mathrm{\Omega }$ resistance is now connected in parallel to the resistance in the right gap, isCase 1

⇒R4=11$\Rightarrow {\displaystyle \frac{R}{4}}={\displaystyle \frac{1}{1}}$
R=4$R=4$ 
⇒42=x100−x$\Rightarrow {\displaystyle \frac{4}{2}}={\displaystyle \frac{x}{100-x}}$
200=3x$200=3x$ 
x=2003$x={\displaystyle \frac{200}{3}}$
Shift =2003−50=503=16.7$={\displaystyle \frac{200}{3}}-50={\displaystyle \frac{50}{3}}=16.7$ cm

Definition

Electrical power

Electrical power is given by: P=H/t=VI=I2R$P=H/t=VI=I^{2}R$
Its unit is W$W$ (Watt).

Definition

Sign convention of electrical power

The power will be a positive if energy is leaving the circuit (energy is absorbed) by way of the element. In this case, energy is being converted to a non-electrical form, such as heat, light, motion, chemical energy, an electric field, or a magnetic field.

The power will be negative if energy is entering the circuit (energy is supplied). Energy is being converted from a non-electrical form to electrical current.

Example

Power dissipated in resistors

Three equal resistors connected in series across a source of emf together dissipate 10 W$10W$ power. The power dissipated if the same resistors are connected in parallel is:Case 1
P=V23R=10$P={\displaystyle \frac{V^2}{3R}}=10$ Watt
Case 2
P=3V2R$P={\displaystyle \frac{{3V}^2}{R}}$
=9(V23R)$=9\left({\displaystyle \frac{V^2}{3R}}\right)$
=90 W$=90W$

Definition

Power supplied by a cell

Power supplied by a cell in a circuit is given as 
P=VI=V2r+R$P=VI={\displaystyle \frac{V^2}{r+R}}$
where V$V$ is the emf of the cell, r$r$ is the internal resistance and R$R$ is the load of the circuit.

Example

Energy consumption of common electrical appliances

An electric heater of resistance  6 Ω$6\mathrm{\Omega }$ is run for 10$10$ minutes on a line of 120 volt$120volt$. The energy liberated in this period of time is:E=V2Rt$E={\displaystyle \frac{V^2}{R}}t$

=120×1206×10×60$={\displaystyle \frac{120\times 120}{6}}\times 10\times 60$

E=14.4×105J$E=14.4\times {10}^{5}J$

Definition

Maximum power transfer theorem

The maximum power transfer theorem states that, to obtain maximum external power from a source with a finite internal resistance, the resistance of the load must equal the resistance of the source as viewed from its output terminals.
(PR)max=ϵ24r$(P_R{)}_{max}={\displaystyle \frac{{ϵ}^2}{4r}}$
when R=r where ϵ$ϵ$ is the emf of the cell and r is the internal resistance.
Example : Two identical batteries each of e.m.f. 2 V and internal resistance 1 Ω$\mathrm{\Omega }$ are available to produce heat in an external resistance by passing a current through it. The maximum power that can be developed across R using these batteries isFor maximum current, the two batteries should be connected in series. The current will be maximum when external resistance is equal to the total internal resistance of cells i.e R=r+r=1+1=2Ω$R=r+r=1+1=2\mathrm{\Omega }$
Thus, E+E=I(R+r+r)⇒I=2ER+2r=2×22+2=1A$E+E=I(R+r+r)\Rightarrow I=\frac{2E}{R+2r}=\frac{2\times 2}{2+2}=1A$
Maximum power, P=I2R=12×2=2W

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