2018-10-01_Physics_For_You

zero and that due to capacitance

1

ωC









becomes

innite, so equivalent circuit is shown in gure. R/ 2

R/ 2

R

K

EE

r 1 r 2

Net external resistance R

R

ext= R

+

(^2) =
2

3

4

Net internal resistance Rint = r 1 + r 2

? Current in circuit I

E

Rrr

=

++

2

3

4 12

e potential dierence across the terminals of cell A is zero; so E – Ir 1 = 0

E

Er

Rrr

− Rrr ++

=⇒=−

2

3

4

0

4

3

1

12

() 12

(c) : Current in branches containing L and C will
ow independently

K

I

I 1 I 2

L C

10 : 10 : 20 V

Ie

t

1 510

20

10

1

3

2

15

4 =−







==

− × − .A

Ie

t

2 10

20

10

3 ==

− − .A ? I = I 1 + I 2 = 2.5 A

(a) : Inductive reactance,
XL = ZL = 2p × 50 × 35 × 10–3

=× 2 ×××=− Ω

22

7

50 35 10^311

erefore, impedance of circuit ZR=+^22 XL=+ 112211 =Ω 11 2 Given Vrms = 220 V Amplitude of voltage VV 0 ==^2 rms^2202 V

Amplitude of current i

V

(^0) Z

0 220 2

11 2

== = 20 A

Phase lag of current over voltage

φ

ωπ ==tant−−^11 an ==tan(−^1 )

11

1

4

L

R

(b) : From the rating of the bulb, the resistance of
the bulb is

R

V

P

==Ω

2 100

For the bulb to be operated at its rated value, the rms current through it should be 1A.

Also, I

V

rms Z = rms

? 1

200

100 250

3

22

=

+

⇒=

()π L π

L H

(a)

(c) : V = V 0 sin Zt

If it is applied across an inductor, then I lag by

π 2

.

us, I = I 0 sin ω

π t−







2 

P = VI = V 0 I 0 sin Zt sin ω

π  t− 



2 

= −







VI

(^00) tt
2
2sinωωcos

P = −









VI 00

2

sin (2Zt) =

VI 00

2

sin (2Zt + p)

P has an angular frequency of ‘2Z’.

(c)

1A

200 V, 50 Hz

L R = 100 :

COMICCAPSULE

How do you keep warm in a cold room?

You go to the corner, because it’s always 90 degrees.