MPS/PHY-XII-2011/A10
PHYSICS
ASSIGNMENT-10
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1.
The instantaneous voltage from an a.c. source is given by E = 300 sin 314t. What is the r.m.s
voltage of the source?
Comparing the expression: ε=300sin (314t) with ε=ε0 sin( 2πft), we find that,
Peak voltage, ε0=300V and 2πf = 314
Therefore frequency f = 314/2π = 50Hz
2.
Can we use a.c. of frequency 15cps for lighting purpose?
Yes, because the fluctuations in current will be so rapid (30 times per second) that the
bulb will appear glowing continuously due to persistence of vision.
3.
Find the time required for a 60 Hz alternating current to reach its peak value starting from
zero.
Time period of a.c T= 1/f = 1/60 s.
The current will take one-fourth of the time period to reach its peak value starting from
zero.
Therefore required time t = T/4 = 1/240 s
4.
The divisions marked on the scale of an a.c. ammeter are not equally spaced. Why?
An a.c ammeter is based on heating effect of current. As the heat produced varies as the
square of current (not directly with current), so the divisions marked on the scale are
not equally spaced.
5.
Capacitors block d.c. Why?
For d.c , f= 0 therefore, Xc = 1/2πfC = ∞.
So a capacitor does not allow d.c. to flow through it i.e., it blocks d.c.
6.
Show that an inductor offers an easy path to d.c. and a resistive path to a.c.
Inductive reactance, XL = 2πFL
Clearly XL is zero for d.c. (f=0) and has a finite value of a.c. (finite f). Hence an inductor
offers an easy path to d.c. and a resistive path to a.c.
7.
Sketch graphs showing the variation of (i) resistance (ii) inductive reactance (iii) capacitive
reactance with frequency of the applied voltage.
8.
What is the phase difference between the voltages across L and C in a series LCR-circuit
connected to an a.c. source?
Given a current in series LC, voltage in L leads current by 900 phase and voltage in C lags
behind current by 900 phase. So voltages in L and C differ by a phase of 1800
9.
A choke coil and a bulb are connected in series to an a.c source. The bulb shines brightly. How
does its brightness change when an iron core is inserted in the choke coil?
When the iron core is inserted in the choke coil, the self-inductance L increases.
Consequently, the inductive reactance, XL = ωL increases. This decreases the current in
the circuit and the bulb glows dimmer.
10.
An inductor ‘L’ of reactance XL, is connected in series with a bulb ‘B’ to an a.c. source as
shown. Briefly explain how does the brightness of the bulb change, when (i) number of turns of
the inductor is reduced and (ii) a capacitor of reactance XL = XC is included in series in the
same circuit.
(i)
When the number of turns in the inductor is reduced, its reactance XL
decreases. The current in the circuit increases and hence brightness of the bulb
increases.
(ii)
With capacitor of reactance XC = XL, the impedance
Z = R 2 + ( X L − X C )2 = R
becomes minimum. The current in the circuit becomes maximum. The bulb
glows with maximum brightness.
11.
In a series R-C circuit, R=30Ω, C = 0.25µF, V = 100V and ω = 10000 rad/s. Find the current in
the circuit and calculate the voltage across the resistor and the capacitor. Is the algebraic sum
of these voltages more than the source voltage? If yes, resolve the paradox.
Here R = 30Ω, C = 0.25 X 10-6F
Vrms=100V, ω = 10,000 rad/s
XC= 1/ωC = 400Ω
Z = ( R2+ XC2)1/2 = 401.1Ω
Irms = Vrms/Z = 0.25 A
Vrms(R) = IrmsR = 7.5 V
Vrms(C) = IrmsXC = 100 V
12.
An a.c circuit consists of a series combination of circuit elements ‘X’ and ‘Y’. The current is
ahead of the voltage in phase by π/4. If element X is a pure resistor of 100Ω, (i) name the
circuit element ‘Y’ and (ii) calculate the rms value of current, if rms value of voltage is 141 V.
(i)
The circuit element ‘Y’ is a capacitor
(ii)
Phase angle , φ = π/4
But cos φ = R/Z Therefore cos (π/4)=100Ω/Z
X = 100/(1/√2) =141.4Ω
Irms = Vrms/Z = 1 A
13.
A 100 µF capacitor in series with a 40 Ω resistance is connected to a 110 V, 60 Hz supply.
(a) what is the maximum current in the circuit? (b) What is the time lag between current
maximum and voltage maximum?
V0
For a RC circuit, if V = V0cosωt, then i =
1
R + 2 2
ωC
cos(ωt + φ ) where tanφ = 1/ωCR
2
Here Veff = 100V ,ω=2πf = 2π × 60 rad/s , R = 40Ω, C = 100µF = 10-4F
(a) Maximum current in the circuit is
i0 =
V0
2Veff
=
1
R + 2
ω + C2
2
=3.24 A
1
R + 2
ω + C2
2
(b) The phase angle φ is given by tanφ = 1/ωCR =0.6631
Therefore φ = 33.50
Time lag, ∆= φ/ω = 1.5 ×10-3 = 1.55 ms
Here the voltage lags behind the current or the current leads the voltage.
14.
What happens to answers (a) and (b) in Question 13. if the circuit is connected to a 110 V,
12kHz supply. Hence, explain the statement that a capacitor is a conductor at very high
frequencies. Compare this behavior with that of capacitor in a d.c. circuit after the steady state.
Here R = 40Ω, C = 100µF = 10-4F
Erms = 110 V, f= 12kHz = 12 × 103Hz
(a) XC = 1/2πfC = 0.133Ω
irms =
ε rms
R 2 + X C2
= 2.75 A
therefore io= √2 irms = 3.89 A
(b) tan φ = XC/R = 0.0033 or φ= 0.20 ≈ 00
Now in the absence of capacitor, irms = εrms/R = 2.75 A
Hence at very high frequency (12 kHZ), the current in the circuit is same both in the
presence or absence of the capacitor. It follows that at high frequency, capacitor acts
like a conductor. But for a d.c supply, f= 0 => Xc = ∞ i.e, capacitor blocks d.c.