Download Module-I , Leture-10

Module-I , Leture-10
Numerical examples on DC Circuits and Magnetism
# A current of 20A flows through two ammeters A and B in series. The potential
difference
across A is 0.2V and across B is 0.3V. Find how the same current will divide between
A and B when they are in parallel. [Dec 2014/Jan2015] Case I:
V1 = 0.2 V
V2 = 0.3 V
Resistance of ammeter A, R1 = V1/I = 0.2/20 = 0.01 ohms
Resistance of ammeter B, R2 = V2/I = 0.3/20 = 0.015 ohms
Case II:
I1 = I* (R2/ (R1+R2)) = 20*(0.015/ (0.015+0.01)) = 12A
I2 = I* (R1/ (R1+R2)) = 20*(0.01/ (0.015+0.01)) = 8A
# Coils A and B in a magnetic circuit have 600 turns and 500 turns respectively. A
current of 8A in coil A produces a flux of 0.04Wb. If co-efficient of coupling is 0.2,
calculate i) self-inductance of the coil A with B open circuited, (ii) flux linking with
the coil B (iii) the average emf induced in coil B when flux with it changes from zero
to full value in 0.02
seconds, (iv) mutual inductance.
2014]
[Dec 2014/Jan2015, Dec 2013/Jan
Solution:
NA = 600, NB = 500, IA = 8A, ФA = 0.04 Wb, K = 0.2
1.
2.
LA = =
=
= 3H
K = 0.2
= 0.008H
in coil B, = -NB
3.
4.
M==
= - 500 ×
= -500 ×
= -200 V
= 0.5H
# A circuit consists of 2 parallel resistors having resistances 20Ω and 30Ω respectively,
connected in series with a 15Ω resistor. If the current through 30Ω resistor is 1.2A, Find
(i) Currents in 20Ω and 15Ω resistors (ii) The voltage across the whole circuit (iii)
voltage
across 15 Ω resistor and 20 Ω resistor (iv) total power consumed in the circuit.
[Dec
2014/Jan2015]
Solution : Voltage across 30 Ω is = I R = 1.2 × 30 = 36V
Current in 20 Ω is = V/R = 36/ 20 = 1.8 A (Since 20 Ω and 30 Ω are in parallel)
Total current in the circuit is = 1.8 +1.2 = 3 A
Voltage in 15 Ω is = 3 ×15 = 45 V
Total voltage is = 45 +36 = 81 V
Power in the circuit is = VI = 81 ×3 = 243 W
# A coil consists of 600 turns and a current of 10A in the coil gives rise to a magnetic
flux of 1mWb.Calculate (i) self-inductance (ii) induced emf (iii) energy stored when
the current is
reversed in 0.01second.
2014/Jan2015]
[Dec
L=
= 0.06 H
=
e = -L
Energy stored =
= - 0.06
(-10-10)/0.01 = 120V
= (1/2)* (0.06)* 102 = 3 J
# Two coils having 1000 turns and 1600 turns respectively are placed close to each
other such that 60% of the flux produced by one coil links the other. If a current of
10A, flowing in the first coil produces a flux of 0.5mWb. Find the inductance of the
second coil.
[June/July 2014]
Solution: NA = 1000, NB = 1600, IA = 10A, ФA = 0.5mWb.
LB = ? ФB = 60% of ФA = 0.6× 0.5mWb = 0.3mWb
LB = =
#
= 0.048 H
Find the resistance of the circuit shown (RAD).
[ June/July 2013]
((2 Ω || 5 Ω || 10 Ω) + (6 Ω || 4 Ω) + 1.35) || 5 Ω
(1.25 + 2.4 +1.35) || 5 Ω
= 2.5 Ω
# In the parallel arrangement of resistors shown the current flowing in the 8Ω
resistor is 2.5A. Find current in others resistors, resistor X, the equivalent resistance.
[June/July
2013]
V=IR=2.5*8=20V
I40= =
=0.5A
I25 =
= 0.8A
Ix = 4-(2.5+0.8+0.5) = 0.2A
Therefore, X (Ω) =
=
= 100Ω
# Find the value of resistance R as shown in the figure below. So that the current drawn
from the source is 250 mA. All the resistance are in ohms.
2014]
Req = [(R || 40) + 40] || 30
Req =
Req =
=
R = 40Ω
[Dec 2013/Jan