DC Current

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