A current I flows along the length of an infinitely long, straight, thin walled pipe. Then
the magnetic field is zero only on the axis of the pipe
the magnetic field is different at different points inside the pipe
the magnetic field at any point inside the pipe is zero
the magnetic field at all points inside the pipe is the same, but not zero
C.
the magnetic field at any point inside the pipe is zero
The phase difference between the alternating current and emf is π/2. Which of the following cannot be the constituent of the circuit?
C alone
R.L
L.C
L alone
B.
R.L
0<phase difference for R-L circuit < π/2
In a LCR circuit capacitance is changed from C to 2C. For the resonant frequency to remain unchanged, the inductance should be changed from L to
4L
2L
L/2
L/4
C.
L/2
In the condition of resonance X_{L} = X_{C}
or
ω_{L} =1/ ω_{C} = .................. (i)
Since, resonant frequency remains unchanged
So, = constant
or L_{C} = constant
∴ L_{1} C_{1} = L_{2} C_{2} ⇒ L×C = L2× 2C
L_{2} = L/2
In an LCR series a.c. circuit, the voltage across each of the components, L, C and R is 50 V. The voltage across the LC combination will be
50 V
50√2 V
100 V
0 V(zero)
D.
0 V(zero)
In an LCR series a.c. circuit, the voltage across inductor L leads the current by 900 and the voltage across capacitor C lags behind the current by 900. Hence, the voltage across LC combination will be zero.
In a series resonant LCR circuit, the voltage across R is 100 volts and R = 1 kΩ with C = 2µF. The resonant frequency ω is 200 rad/s. At resonance the voltage across L is
4 × 10−3 V
2.5 × 10^{−2} V
40 V
250 V
D.
250 V
In a series LCR circuit, R = 200 Ω and the voltage and the frequency of the main supply is 220 V and 50 Hz respectively. On taking out the capacitance from the circuit the current lags behind the voltage by 30º. On taking out the inductor from the circuit the current leads the voltage by 30º. The power dissipated in the LCR circuit is
305 W
210 W
zero
242 W
D.
242 W
The given circuit is under resonance as X_{L} = X_{C. }Hence, power dissipated in the circuit is
P = V^{2}/R = 242 W
Two voltameters one of copper and another of silver, are joined in parallel. When a total charge q flows through the voltameters, equal amount of metals are deposited. If the electrochemical equivalents of copper and silver are z_{1} and z_{2} respectively the charge which flows through the silver voltameter is
B.
Alternating current can not be measured by D.C. ammeter because
A.C. cannot pass through D.C.
A.C. changes direction
average value of current for complete cycle is zero
D.C. ammeter will get damaged.
C.
average value of current for complete cycle is zero
The full cycle of alternating current consists of two half cycles. For one half, currently is positive and for second half, current is negative. Therefore, for an a.c. cycle, the net value of current average out to zero. While for the half cycle, the value of current is different at different points. Hence, the alternating current cannot be measured by D.C. ammeter.
A circuit has a resistance of 12 Ω and an impedance of 15 Ω. The power factor of the circuit will be
0.8
0.4
1.25
0.125
A.
0.8
In an a.c. circuit the voltage applied is E = E_{0} sinπt. The resulting current in the circuit is I = I_{0} sin .The power consumption in the circuit is given by
P = zero
B.
P = zero