Simple d.c. Circuits 3

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Simple d.c. Circuits

  At the end of this lesson, the students should be able to:

• Solve problems involving resistivity and conductivity
 • Deduce the equivalent resistance of series and parallel circuits
 • Solve problems on series and parallel resistors


 2.7  SOLVE PROBLEMS INVOLVING RESISTIVITY AND CONDUCTIVITY

Example 7.1 
A cylinder wire of 0.5m in length, 0.5mm in diameter, has a resistance of 2.5Ohms. calculate the  resistivity of the wire.

Solution

Example 7.2

A wire of diameter 0.6mm, resistivity 1.1 X 10^-6Ω - m has resistance of 44Ω. Calculate the

length of the wire. 

Solution Here, we are given: 
radius, r = 0.6 mm = 3 X 10^-4m
                       2
To find l  we 
                                                  =11.3m

Example 7.3
A 30m conductor of cross sectional area 2mm^2 has a resistance of 10Ω. Calculate the
conductivity of the conductor.

Solution



2.8 EQUIVALENT RESISTANCE OF SERIES AND PARALLEL CIRCUIT 
Equivalent Resistance Of A series Circuit


When resistors of resistance R1, R2 and R3 are joint together (fig 7.1), they are said to be connected in series. It should be noted that:

(i) The current through a series circuit is the same all over.
(ii) The voltage drop across each resistor is different depending on its resistances.
(iii) The applied voltage (Vs) is equal to the sum of the P.d’s across each resistor.

     i.e. I=I1=I2=I3
Vs= V(R1) + V(R2) +V(R3)
(7.1)
where Vs = IR; V(R1)= IR1; V(R2) = IR2; V(R3) = IR3
(7.2) 

Putting eqtn (7.2) in (7.1) gives
IR = IR1 + IR2 + IR3
=> Req = R1 + R2 +R3

For n resistors connected in series, the equivalent resistance

Req. is given by Req. = R1 + R2 +R3 + …. + Rn
(7.3)

Equivalent Resistance Of A Parallel Circuit


When resistors of resistance R1, R2 and R3 are connected in parallel, the following statements hold good:

(i) The p.d across each resistor is the same
(ii) The current flowing through each resistor is different
(iii) The supply current (I) is equal to the sum of current flowing through each resistor

i.e. Vs = V1 = V2 = V3 I = I1 + I2 + I3
(7.4)
where I = Vs/R; I1 = Vs/R1 ; I2 = Vs/R2; I3 = Vs/R3
(7.5)

 Inserting eqtn (7.5) in (7.4) gives Vs/R = Vs/R1 + Vs/R2 + Vs/R3
⇒ 1/R = 1/Req = 1/R1 + 1/R2 + 1/R3 

For n resistors connected in parallel, 1/Req = 1/R1 + 1/R2 + 1/R3 + - - - + 1/Rn
(7.6)







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