Question

can someone please help me with these problems

Answer 1

hello friend,

i). Given:

L1 = 1m. then R1 = 6 Ohm.

L2 = 70cm. = 70/100
= 0.7m then R2= ??

1m = 6 Ohm
0.7m = ???

therefore,

l1/l2 = R1/R2

1 / 0.7 = 6/R2

: R2 = 0.7*6/1

: R2 = 4.2 Ohm <Ans

--------------------------------------------------

ii).

Let x and y be the two resistance

Given,

x + y = 80. when they connected in series combination.

1/x + 1/y = 1/20 = 20 when they are connected in parallel combination.

: x + y/xy = 1/20

: 80/xy = 1/20

80*20 = xy

1600 = xy

1600/y = x

1600+y/y =80

1600+y^2 = 80y

y^2-80y+1600= 0

y^2-40y-40y+1600=0

y(y-40)-40(y-40)= 0

y= 40..

now we have to find the value of x.

x+y = 80

x+ 40= 80

x= 80-40

X = 40.

hence the value of two resistance of x and y are 40 Ohm and 40 Ohm.

---------------------------------------------------

iii). Given.

Q = 420 C
t = 5 minute= 5*60= 300 second

then

I= Q/t

I = 420/300

I = 42/30. ......zero zero cancelled.

I = 1.4 A. ( Divided)

..........I hope it will helps you........

@@@@@@@@@@@@@@@@@@@@

@@@#$#$##. Raju Singh #$#$#$@@@