Question

A resistance of 20 ohm, an inductance of 0.2 H and a capacitance of 100 uF are
connected in series across 220-V, 50-Hz mains. Determine the following (a) impedance
(b) current (c) voltage across R, L and C (d) power in watts and VA (e) p.f. and angle of
lag.​

Answer 1

Explanation:

XC = 0.2 × 314 = 63Ω , C = 10μF = 100 × 10−6 = 10−4 farad XC = 1/ωC = 1/(314 x 10-4) = 32Ω, X = 63 − 32 = 31 Ω (inductive) (a) Z = √(202 + 31)2 = 37 Ω (b) I = 220/37 = 6 A (approx) (c) VR = I × R = 6 × 20 = 120 V; VL = 6 × 63 = 278 V, VC = 6 × 32 = 192 V (d) Power in VA = 6 × 220 = 1320 Power in watts = 6 × 220 × 0.54 = 713 W (e) p.f. = cos φ = R/Z = 20/37 = 0.54; φ = cos−1(0.54) = 57°18′

Answer 2

Given:

Resistance (R) = 20Ω

Inductance (L) = 0.2 H

Capacitance (C) = 100μF

Voltage (V) = 220 V

Frequency (f) = 50 Hz

To find:

1. Impedance (Z)
2. Current (I)
3. Voltage across L, C, R
4. Power (P)
5. Power factor (∅) and angle of lag

Solution 1

We have been given the value of L, C and R and also the frequency

Hence,

Inductive reactance ($X_{L}$) = 2π f L

Substituting the given values, we get

$X_{L}=2(\frac{22}{7} )(50)(0.2)$   $;$ $or$

$X_{L}$ = 62.8 Ω

Similarly, we have

Capacitive Reactance $X_{C}$ = (2π f C)⁻¹

Substituting the values, we get

$X_{C}=\frac{10^{6} }{2(3.14)(50)(100)}$   $;$ $or$

$X_{C}$ = 31.8 Ω

We know,

Impedance $Z=\sqrt{R^{2}+ (X_{L}- X_{C} )^{2} }$

Substituting the given values, we get

$Z=\sqrt{(20)^{2}+(62.8-31.8)^{2} }$  

$Z=\sqrt{400+961}$

$Z=\sqrt{1361}$   $;$ $or$

$Z=36.8$ Ω

Hence, the Impedance of the circuit is 36.8 Ω.

Solution 2

From Ohm's Law, we know, $I=\frac{V}{R}$

Here, $R=Z$

Therefore,

$I=\frac{220}{36.8}$   $;$ $or$

$I = 6A$

Hence, the current flowing through the circuit is 6 A.

Solution 3

Similarly, applying Ohm's law, $V=IR$ on individual devices, we get

$V_{R}= 6(20) =120V$

$V_{L}=6(62.8)=376.8V$

$V_{C}=6(31.8 ) =190.8V$

Solution 4

We know, Power $P=VI$

Substituting the given values, we get

$P=220(6)$    $;$ $or$

$P=1320W$

Hence, the power in the circuit is 1320 Watts.

Solution 5

Power factor (∅) = cos ∅

And, we know,

cos ∅ = $\frac{R}{Z}$

Therefore, substituting the known values, we get,

cos ∅ = $\frac{20}{36.8}$

Hence, power factor cos ∅ = 0.5

And,

Angle of lag (∅) = $cos^{-1}(0.5)$ = $60^{o}$

Final answer:

Hence,

1. Impedance of the circuit is 36.8Ω.
2. Current in the circuit is 6 A.
3. Voltage across R is 120 V , across L is 376.8Ω and across C is 190.8 V.
4. Power of the circuit is 1320 watts.
5. Power factor of the circuit is 0.5 and the angle of lag is 60°.