Question

A dry cell of EMF 1.5 volt and internal resistance 0.10 ohm is connected across a resistor in series with a very low resistance ammeter.when the circuit is swithched on the ammeter reading settles to steady value of 2A.what is the
(a) rate of chemical change consumption of the cell
(b) rate of energy dissipation inside the cell.
(c) rate of energy dessipation inside a resistor.

Answer 1

@shruti you here ?

Answer 2

yes

Answer 3

http://www.ausetute.com.au/ratelaw.html
try the link for the first part i m kinda missed you right now.

Answer 4

thanks

Answer 5

@annas link is of chem.
and my question is of physics.

Answer 6

ok i got it
E*I = rate of chemical change consumption of the cell

Answer 7

ok
we have
I=2A
V=1.5 volt
R=0.10 ohm
for the parts C
energy dassipation inside resistor is given by

P=I^2R=(2)^2(0.10)=0.4 watt

Answer 8

1.5 * 2= 3 watt

Answer 9

second part is done by sami

Answer 10

3) I= E/(R+r)

Answer 11

now for the
part a
rate of consumption should be equal to power output (
p=IE=1.5*2=3W

Answer 12

2= 1.5/R+0.10

Answer 13

R= 0.65 ohm

Answer 14

@shruti the link was pretty similar to the question i know it was chemistry related i thought it would help :)

Answer 15

okay,
let me check it once.

Answer 16

< signifies?

Answer 17

as anas found R=0.65
so
rate of energy dessipation inside a resistor is

P=I^2R=(2)^2*0.65=2.6 Watt

Answer 18

@sami-21 you're genius man...

Answer 19

what does "<" signifies?

Answer 20

'<' is a sign of inequality i've no idea what it signifies here

Answer 21

'<'it means less than where it is mentioned here?

Answer 22

P=I^2R=(2)^2*0.65=2.6 Watt

Answer 23

"^" is for power

Answer 24

it means \[p=i^2R\]
it means 2 is in the power of I

Answer 25

exponent