**Electric field II: Current Electricity**

**(A) Electric current**

A current is any motion of charge from one point to another in conducting materials. Current is donated by *I*, therefore current *I* through the cross sectional area, A, to be the net charge, Q, flowing through the area per unit time. That is *I* = Q/t(Cs^{-1}) or ampere (A).

**(B)Electromotive force and circuits**

*We defined Electromotive force (e.m.f) as the magnitude of the potential difference of both the external circuit and the inside of the cell. The SI unit is volt (V) for steady current circuits, examples of sources of e.m.f. are batteries, d.c. electric generator, solar cells, thermocouples and fuel cells. Such devices convert energy of some form (mechanical, chemical, thermal and so on) into electrical potential energy and transfer it into the circuit to which it is connected.*

**Note:** **the potential difference (p.d.) between any points in a circuit is the work done when one coulomb of charge moves from one point to another. **P.d. across external circuit and internal resistor) E= IR + It (IR = p.d. across external circuits and Ir = p.d. across the internal circuit.

Symbols of some electrical components

Source: www.commons.wikimedia.com

**Resistivity and resistance:** the resistivity p(rho) of a material is the ratio of the magnitude of Electric field E and current density J (current per unit area) [p=E/J]. It is also defined in terms of the resistance R of a wire of a unit length, *l*, and a unit cross-sectional area *A*.

** P=AR/ t ~ R= PL**/

*A**Conductivity is given by the expression*

* 1/p = ∆*

* = 1/RA*

*In Siemens per meter(Sm ^{-1})*

*The resistivity of a metallic conductor that increases with increase in temperature.*

* p*_{1}* = p*_{0}*[1+∆(t-t*_{0}*)] where p*_{0}* is the resistivity at temperature, t*_{0}* and ∆ is the coefficient of resistivity.*

**(C)Ohm’s law**

*The Electric current flowing through a metallic conductor (wire) at constant temperature is directly proportional to the potential difference between its ends. Ohm’s law can expressed by the equation as: V = kL*

*The constant of proportionality, k is the resistance, R of the wire such that **V=IR*

*This relationship is Ohm’s law provided R is a constant.*

**Note:** internal resistance donated by r.

From ohm’s law. The e.m.f E = V+lr

= IR+Ir

so, the current, *l,* through the external circuit and connected to the source is

* I = E/ R+r*

**(D)Resistors in series**

Resistors in series

Source: www.commons.wikimedia.org

Combined resistance R= V/I

Where *R= R*_{1}* + R*_{2}* + R*_{3}

Rheostats

Source: www.needpix.com

**Rheostats** consist of a long resistance coiled round on an insulator. The resistance introduced in the circuit depends on the length of the coil as determined by the slide position.

**Resistors in parallel**

Resistors in parallel

Source: www.commons.wikimedia.org

Resistors in parallel can be expressed as

*1/R = 1/R*_{1}* + 1/R*_{2}* + 1/R*_{3}

*For two resistances R*_{1}*, and R*_{2}*, the equivalent resistance R*_{12}* = **R*_{1}*R*_{2}* / R*_{1}* + R*_{2}

**(E) Measurement of resistance by the ammeter voltmeter method**

Determination of resistance of a resistor

Source: www.en.wikipedia.org

**Note** that a resistor of known resistance, R, can be constructed from a length, *l*, of manganin wire of known resistivity,p.

*l* = Rπd^{2}/ 4p

Wheatstone bridge

Source: www.commons.wikimedia.com

This is “balance” condition and

*R*_{1}* / R*_{2}* = R*_{3}* / R*_{4}

Metre bridge

Source: www.commons.wikimedia.com

**Note:** the value of resistance

= X/R = l / 100-l

**Precautions taken to obtain accurate results**

- The resistance R should be selected so that the balance point C comes fairly near to the centre of the wire.
- Very small current should be passed through the galvanometer to avoid damage. A protective resistor should be connected to the galvanometer.
- Tight connection should be ensured.
- The Jockey should not be dragged along the wire.

**The Potentiometer: **consists of a uniform resistance wire of length 100cm. Since R is the resistance per centimeter, then: R~l

And since the current is constant, then

V_{AC} = kl. Where k is the constant.

Electronic Potentiometer

Source: www.commons.wikimedia.com

**Comparison of e.m.fs. of two cells**

** ***E*_{1}* / E*_{2}* = l*_{1}* / l*_{2}

**Accuracy**

- When the potentiometer is used to compare the e.m.fs. of cells, no errors are introduced by internal resistances since no current flows at balance.
- It is more accurate than the moving coil voltmeter for measuring e.m.f.
- For a very sensitive galvanometer, a high precision of the balance point of the potentiometer is possible.

**(F)Kirchhoff’s law**

*Definitions of terms*

*(I) a junction (node) in a circuit or electrical network is a point where three or more connecting wires meet.*

*(II) a loop is any closed conducting path*

*Kirchhoff’s law consists of the following:*

*Kirchhoff’s current law (KIL)*

*The algebraic sum of the currents flowing into any junction N is zero. The sum of the currents flowing towards a junction, N is equal to the sum of the currents flowing away from the junction. E I = 0*

*Kirchhoff’s voltage law (KVL)*

*The algebraic sum of the potential difference in any closed loop (including those associated with e.m.f’s and those of resistive elements) must equal to zero. The algebraic sum of the e.m.fs. in any closed loop of an electrical network is equal to the algebraic sum of the voltage (IR) drops in that loop. EE – EIR = 0.*

*(G)Instrument used for commercial purposes*

Voltmeter and ammeter devices

Source: www.freesvg.org

Conversion of a milliammeter to ammeter and voltmeter respectively

**Questions**

1.A wire of 0.50m long and of diameter 0.80mm has a resistance of 2.5ohms . Calculate the resistivity of material of which it is made and (ii) determine its conductivity.

**Solutions**

R= pl/A

Where R = 2.5, l=0.5m, A=π×0.64×10^{-6}/ 4

(I) resistivity p = RA/1 = 22/7 × 0.64×10^{-6}/5×10^{-1}

= 1.0×10^{-5}ohm/m

(II) conductivity ∆= 1/p = 10^{5}ohm/m

2. A voltmeter of 800ohm resistance reads 180V when connected across a battery of 30ohm internal resistance. Calculate the emf of the battery.

**Solution:** V=IR

I= 180/800 = 9/40 A

Therefore, E= I(R+r)

= 9/40 (830) = 186.75V

3. Which of the following instruments is not suitable for measuring current?

A. Ammeter

B. Voltammeter

C. Milliammeter

D. Microammeter

**Answer:** D

4. Define current?

**Solution:** current can be defined as any motion of charge from one point to another in conducting materials.

## No Comments