What is the reading shown by the ammeter?

[leena19]

Homework Statement

3)An ammeter has a full scale deflection of 0.5 A.Internal resistance is 12 ohms.Explain with necessary calculations,how it should be converted to an ammeter of full scale deflection 2A.
4) does this modification change the sensitivity of the instrument?explain
5. The modified ammeter is connected to the circuit given in the diagram to measure the current through the appliance of resistance 1 ohm
a. what is the current through the appliance?
b. what is the reading shown by the ammeter?
c. calculate the potential difference between points C & F
http://img199.imageshack.us/img199/793/ammeter.png

Homework Equations

Kirchoff's laws

The Attempt at a Solution

1) T find the shunt resistance(R),
0.5*12 = 1.5*R
R=4 ohms

2) sensitivity = deflection(\theta per unit current(I)
so I = C(torsion constant of springs)/BAN\theta
so sensitivity=BAN/C,
looks as though the current has no effect on the sensitivity ,so I guess the answer is no?

3) a.) since the current through the ammeter is 0.5A,and the current throgh the shunt is 1.5A,the current through the appliance,should be 2A?

b) 0.5A?
c) Is it correct to take the potentials,
V_DE = V_CF &
(2*1) +( 0.5*12)= 8V
I'm very very doubtful of this part,cause if I do it like this ,I won't be using any of the other values given in the circuit,and it looks a bit odd?

THANK YOU

 

[rl.bhat]

You have to use Kirchhoff law to find the current through the ammeter. While doing so you have to take into account the combined resistance of the ammeter and the shunt.

 

[leena19]

Okay,Thanks.so using kirchhoff laws,
taking the current through the 4 V cell as, I,
current through CF = I₂
current through DE = I₁,

\stackrel{\rightarrow}{ABCFGA}
I*1 + 2I +I₂ + 3I₂ = 6
3I + 4I₂ = 6 ---------------(1)

combined resistance in ammeter = 3 ohms

ABCDEFGA,
I + 2I +I₁ + 3I₁ = 4
3I +4I₁ = 4 ---------------------(2)

CDEFC,
4I₁ - 3I₂ = -2 ----------------------(3)

(2)-(3),
3I + 3I₂ = 6----------(4)

(1) =(4)
3I + 4I₂ = 3I + 3I₂
i GET I₂ = 0, Is this possible?

4I₁ = -2
I₁=-0.5A
so the current through the appliance is -0.5A?

3)b)
i=current through the ammeter
-0.5-i through the shunt

i*12 = (-0.5-i)*4
i=-0.125A through the ammeter?

3)c)
since there is no current ,I2, through CF, is the potential 0?

 

[rl.bhat]

CDEFC,
4I1 - 3I2 = -2 ----------------------(3)
It should be
CDEFC,
4I1 - 4I2 = -2 ----------------------(3)

 

[leena19]

Yes,i think i missed the 1 ohm resistance due to the 2 V cell,and now i have to redo the whole thing
so,
CDEFC,
4I₁ - 4I₂ = -2 ----------------------(3)

I₁+I₂ = I,
4(I-I₂)-4I₂ = -2
2I - 4I2 = -1 ------------(3)''

(1)+(3)'',
5I=5
I=1A

I1=0.25A which is the current through the appliance

b) I₂ = 1-0.25A = 0.75A

i*12 = (0.25-i)4
i=0.0625A

C) Vc - I₂*1 + 2 - I₂*3 =Vf
Vc-0.25+2-0.75=VF
V_CF= -1V

I hope it's correct now?

 

[rl.bhat]

It seems correct.