A filament bulb (500 W,100 V) is to be used in a 230 V main supply. When a resistance R is connected in series, it works perfectly and the bulb consumes 500 W. The value of R is
(a) 230 (b) 46
(c) 26 (d) 13
(c) If a rated voltage and power are given,
Current in the bulb,
Resistance of bulb,
Resistance R is connected in series.
Which of the following combinations should be selected for better tuning of an L-C-R circuit used for communication ?
(c) For better tuning, peak of current growth must be sharp. This is ensured by a high value of quality factor Q.
Now, quality factor is given by Q=
From the given options highest value of Q is associated with , L=3.5H and C=30
The potential differences across the resistance, capacitance and inductance are 80V, 40Vand 100V respectively in an L-C-R circuit. The power factor of this circuit is
(a) 0.4 (b) 0.5
(c) 0.8 (d) 1.0
(c) Power factor of teh L-C-R circuit
A 100 resistanceand a capacitor of 100 reactance are connected in series across a 220 V source. When the capacitor is 50% charged, the peak value of the displacement current is
(a) 2.2 A (b) 11A
(c) 4.4A (d) 11A
(a) Impedence of the R-C circuit, Z=
where, R= 100 and =100
Peak value of the current,
A small signal voltage V(t)=Vo sinωt is applied across an ideal capacitor C
(a) over a full cycle the capacitor C does not consume any energy from the voltage source
(b) current I(t) is in phase with voltage V(t)
(c) current I(t) leads voltage V(t) by 180°
(d) current I(t) lags voltage V(t) by 90°
(a) For an AC circuit containing capacitor only, the phase difference between current and voltage will be π /2 (i.e. 90°).
In this case current is ahead of voltage by π /2.Hence,power in this case is given by
P=VI cosφ (φ= phase difference between voltage and current)
P=VI cos 90°=0
An inductor 20 mH, a capacitor 50μF and a resistor 40Ω are connected in series across a source of emf V=10sin340t. The power loss in the AC circuit is
(a) 0.67 W (b) 0.78W (c) 0.89 W (d) 0.51 W
L=20mH C=50μf R=40 Ω emf, V=10sin340t
∴ Power loss in AC circuit will be given as
=(10/√2)2 40 [1/402+(340x20x103-1/340x50x10-6)]2
A resistance 'R' draws power 'P' when connected to an AC source. If an inductance is now placed in series with the resistance, such that the impedance of the circuit becomes 'Z' the power drawn will be
When a resistor is connected to an AC source. The power drawn will be
P= Vrms x Irms = Vrms x Vrms/R
When an inductor is connected in series with the resistor, then the power drawn will be
P'=Vrms.Irms cos φ
∴ P'=V2rms/R x R2/Z2=P x R x R/Z2
=>P'=P x R2/Z2=P(R/Z)2
A series R-C circuit is connected to an alternating voltage source.Consider two situations:
1. When capacitor is air filled.
2. When capacitor is mica filled.
Current through resistor is i and voltage across capacitor is V then
Net reactive capacitance
So,current in circuit,I=V/Z
=>I=2πfc/√4π2f2C2R2 +1 x 1/2πfc
i.e. Vc=V/√4π2f2C2R2 + 1
When mica voltage is introduced, capacitance will increase hence, voltage across capacitor get decrease.
A coll of self-inductance L is connected in series with a bulb B and an AC source. Brightness of the bulb decreases when
(a) frequency of the AC source is decreased
(b) number of turns in the coil is reduced
(c) a capacitance of reactance Xc=XL is included in the same circuit
(d) an iron rod is inserted in the coil
(d) As Z=√R2+X2L=√R2+(2πfvL)2
As I=V/Z, P=I2R
i.e., V↑,L↑=>Z↑,I↓ and P↓
In an electrical circuit R, L, C and an AC voltage source are all connected in series. When L is removed from the circuit, the phase difference between the voltage and the current in the circuit is If instead, C is removed from the circuit, the phase difference is again The power factor of the circuit is
Here, phase difference
When, L is removed
When C is removed
Hence, in resonant circuit
It is the condition of resonance therefore phase difference between voltage and current is zero and power factor is cos=1.