DC Current

This document provides information on DC current, including calculations for current, resistance, shunt resistance, and copper resistance. It also includes examples and explanations for solving circuit problems involving current and voltage.

1>DC CURRENT
IM RM a>current i= 200A Imax=1Ma
Is R5> R=0.1Ω Full scale voltage
= I, x R =>1x10-3x0.1 =>0.0001v
Shunt resistance = voltage/ current
= 0.0001/200
= 0.5x10-6Ω
R5nunt=
b>for copper P=1.7x10-8Ωm
cross sectional aria=25cm2
R=P(L/A)
0.5x10-6=1.7x10-8(l/25x10-4)
L = 0.5x10-6x25x10-4/1.7x10-8
=0.7353mm
2>
10-R
9v
10kΩ 5 5kΩ
X
R
R is in parallel with 5kΩ
5R/5+R
Voltage across this combination 3V.and average voltage for loop part is 6v

3(10-R) = 6x 5R/5+R
10-R = 62/3 x 5R/5+R
(10-R)(5+R)= 2X5R
50+10R-5R-R2=10R
R2+5R-50=0
R2 + 10R -5R -50=0
R(R+10)-5(R+10)=0
(R+10)=0 =>R=-10
(R-5)=0 => R=5
3>
I1 i2
E113V R3=3Ω 14v
R=11Ω E2
R2=2Ω
E3 12V
a> Calculate value at current through 12V battery
KCL for top node = I1+I2 ----------------1
=KVL For left hand 100p
=I3 E1 –R3 x I3- E3 –R1 x i1 = 0
=i3-3xi3 i2 – 1x i1=0
=>1-3i3 –i1=0
=>3i3 + i1=1
=>3(i1+i2)+i1=1
=>3i1+3i2+i1=1
=> 4i1+3i2+i1=1-------------------2
=>KVL for right side 100P
=>E2-R3I3-R2-R2Xi2=0
=>14-3xi3-12-2xi2=0
=>2-3i3-2i2=0
3i3+2i2=2
3(i1+i2)+2i2=0
3i1+3i2+2i2=2

3i1+5i2=2----------------------3
3xeqn 2 – 4xeqn3
12i1+9i2=3
12i1+20i2=8
-11i2=-5
I2 = 5/11
=0.4545A
Put that i1 in equation 3
3i1+5x0.4545=2
3i1+2.2725=2
3i1=2-2.2725
I1 =-0.2725/3
=-0.9091amp
I3=i1+i2
=0.45455-0.9091
= 0.36364A
b> powerdedcipated in R1 R2 R3
PR1=I12XR1
=(-0.9091)2X1
=0.82646 Watts
PR2=i22x R2
=(0.45455)2x2
=0.20661 watts
PR3= I32XR3
=(0.36364)2x3
=0.396702 watts
4> No. of turns = 50
Magnetic flux from 10 mwb to 20 mwb
Emf= 62.5 V t=No. of turns x charge in flux/emf
50x(20-10)x10-3/62.5 =>50x10x10-3/62.5
=>8ms
5 B=0.6T
F=1.5N
L=500mm
F=B*I*L
1.5=0.6*I*0.5

I=1.5/0.6*0.5
=5Amp
6 Required radius r of each plate
Capacitance=εrε0A(n-1)/d
Εr=1.00058986 Relative permittivity of air
Ε0=8.8542*10-12 permittivity of vacum (free space)
400*10-12=1*8.8542*10-12*A*(8-1)/(1*10-5)
A=c*d/εrε0
A=400*10-12*1*10-5/7*8.8542*10-12
=0.0645m2
A=1/2 πr2
r=√2*A/π
=√2*64.5376/3.14
=6.41mm
b. i> Initial value of current flowing
I=V/R
=10/50K
=0.2mA
ii>Time constant for the circuit
time constant=R*C
=50*103*400*10-12
=20μs
c.i>Having fully charged, the capacitor is then discharged through the 50 kΩ resistor.
I = (V/R)e^-0.25
= (10/5000)*(0.7788)=1.5576*10^-4=0.1557*10^-3=0.1557mA
ii>the voltage drop across the resistor when the capacitor has been discharging for 10μs

V=10-.5 = 10*(0.60653) = 6.0653volts
7> L =0.2H R=330Ω V= 12vdc
a> Resistance=inductance/time const.
330R=0.2/T
T=0.2/330
=0.606ms
b> Voltage drop across inductor after 2 time constant
I = V/R = 12/330 =>36.36ma
=12 e-(330x0.0012)/0.2
=12 e-1.98 =>1.66V
c> Vtg drop across the inductor after three time constants
=12 e-(330x0.0018)/0.2
=12 e-2.97
= 0.62V
d> Inductor =0.2H
Time constant = 0.55ms
=R=L/T =>0.2/0.55x10-3
= 0.3636kΩ
=963.6Ω