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. [Dec 2014/Jan2015, Dec 2013/Jan 2014]
Solution:
NA = 600, NB = 500, IA = 8A, ФA = 0.04 Wb, K = 0.2
1. LA = = = 3H
2. = K = 0.2 = 0.008H
3. / in coil B, = -NB / = - 500 × / = -500 × / = -200 V4. M = = = 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. [Dec 2014/Jan2015]
L =
= / = 0.06 He = -L = - 0.06 (-10-10)/0.01 = 120V
Energy stored = = (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=20VI40= / = / =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. [Dec 2013/Jan 2014]
Req = [(R || 40) + 40] || 30
Req =
Req =
=
R = 40Ω