1)What gauge of aluminum wire will have the same resistance per unit length as 1

Question

1)What gauge of aluminum wire will have the same resistance per unit length as 13-gauge copper wire?

3)For the circuit shown in the figure, R1=3.00?, R2=4.00?, R3=1.00?, R4=3.00?,R5=1.00?, and the potential difference is 12.0V.

a) What is the equivalent resistance for the circuit?

b) What is the current through R5?

c) What is the potential drop across R3?

4)As shown in the figure, a circuit consists of an emf source with V=19.0V and six resistors. Resistors R1=5.00? and R2=8.00? are connected in series. Resistors R3=6.00?and R4=6.00? are connected in parallel and are in series with R1 and R2. Resistors R5=1.00? and R6=1.00? are connected in parallel and are also in series with R1 and R2.

a) What is the potential drop across each resistor?

?V1=

?V2=

?V3=

?V4=

?V5=

?V6=

b) How much current flows through each resistor?

?V1=

?V2=

?V3=

?V4=

?V5=

?V6=

1)What gauge of aluminum wire will have the same resistance per unit length as 13-gauge copper wire? 2)What is the current in the R1=37.0??resistor in the circuit in the figure? 4)As shown in the figure, a circuit consists of an emf source with V=19.0V and six resistors. Resistors R1=5.00? and R2=8.00? are connected in series. Resistors R3=6.00?and R4=6.00? are connected in parallel and are in series with R1 and R2. Resistors R5=1.00? and R6=1.00? are connected in parallel and are also in series with R1 and R2. a) What is the potential drop across each resistor? ?V1= ?V2= ?V3= ?V4= ?V5= ?V6= b) How much current flows through each resistor? i1= i2= i3= i4= i5= i6= 3)For the circuit shown in the figure, R1=3.00?, R2=4.00?, R3=1.00?, R4=3.00?,R5=1.00?, and the potential difference is 12.0V. a) What is the equivalent resistance for the circuit? b) What is the current through R5? c) What is the potential drop across R3?

Explanation / Answer

1.

Resistance is given by

R=pL/A

Resistance per unit length

R/L=p/A

Diameter of 13 gauge wire is

D=1.828 mm

Area of 13 guage

A=pi*D^2/4=pi*(1.828*10^-3)^2/4 =2.62*10^-6 m^2

Resistance per unit length of Copper wire

(R/L)copper=(1.68*10-8)/(2.62*10-6)=6.4*10-3

Given

(R/L)aluminium=(R/L)copper

paluminium/A=6.4*10-3

(2.82*10-8)/A=6.4*10-3

A=4.41*10-6 m2

pi*D2/4=4.41*10-6

D=2.37 mm

For above Gauge 10 wire has to be used

2.

equivalent resistance

Req=37+(20*20/20+20)=47 ohms

Current flowing through R1 is

I=V/R1=60/47=1.2766 A

3.

R3 and R4 are in series

R_34=1+3 =4 ohms

R2 and R34 are in parallel

1/R_234=1/R_2+1/R_34=1/4+1/4

R_234=2 ohms

R1 ,R234 and R5 are in series ,so equivalent resistance is

Req=2+3+1=6 ohms

b)

Total Current

I5=V/Req=12/6

I5=2 A

c)

Current through R3 and R4 is

I34=I*(4/4+4) =2*(1/2)=1 A

Potential drop across R3 is

V3=I34*R3=1*1=1 Volts

4.

equivalent resistance

R1 and R2 are in series

R_12=5+8=13 ohms

R3 and R4 are in parallel

1/R_34=1/6 +1/6

R_34=3 ohms

R5 and R6 are in parallel

1/R_56=1/1 +1/1

R_56=0.5 ohms

R_12,R_34 and R_56 are in series

Req=13+3+0.5 =16.5 ohms

Total Current flowing in the circuit is

I=V/Req=19/16.5 =1.15 A

So Current through each resistor is

I1=1.15 A

I2=1.15 A

I3=I4=I5=I6=1.15/2 =0.575 A

a)

Potential difference across each resistor

V=I*R

V1=1.15*5=5.75 Volts

V2=1.15*8=9.2 Volts

V3=0.575*6=3.45 Volts

V4=0.575*6=3.45 Volts

V5=0.575*1=0.575 Volts

V6=0.575*1=0.575 Volts

b)

I1=1.15 A

I2=1.15 A

I3=0.575A

I4=0.575 A

I5=0.575 A

I6=0.575 A

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