# thankyou very much sir.these were my 5  questions: Q1 a voltmeter and an ammeter are joined in a series to an ideal cell .giving readings V and A resp.if a resistance equal to resistance of the ammerter is now joined in parallel to the ammeter then  (a) V will not change (b) V will increase © A will become exactly half of its initial value(d)A will become slightly less than double of its initial valueQ2 an ammeter and a voltmeter are connected in series to a batery with an emf E=6volts.whren a certain resistance is connected parallel with the voltmater the reading of the latter decreases two times,wheareas the readinds of the ammeter increase the same no of times.find the voltmere reading after the connection of the resistance. plz explain fully.not able to understand i dont know the principle how to use ammerter and voltmeter resistances to solve circuit problem. if z be any complex no other than 0 then arg(z-i/z+i)= pie/2 then what is the locus of z (not – pie/2,just pie/2) . please explain fully each and every step i am very weak in maths.  plz give procedure and few examples how can i plot the graphs of unusual function(using maxima minima tangents etc...) which are not known from cordinate geometry.i dont understand where to apply graph,where algebra in a maths questions as graph provides answer in very short time .in mathematics AS A WHOLE in which type of questions should i apply graph and where algebra(algebra takes much time).?what is the general procedure for these type of questions-what is the effect of -q in this problem...... if a geometrical figure(eg a cone hemisphere) is given and a point charge q is inside it .then initial flux through curved surface is something(let it be phy.)now another charge -q is placed outside the figure(eg outside the base of the cone).then how to find the net flux through the curved suface in this situation.

Shobhit Varshney IIT Roorkee
8 years ago
Hi,

The answer of first question is as follows:

To make the things simple, assume the reistance of both voltmeter and ammeter be ‘r’.
Before connection of resistance parallel to the ammeter, I = v/2r = 0.5 V/r
V = Ir = 0.5V, A = I = 0.5V/r

After connection, I = V/(3r/2) = 0.67 V/r, V = Ir = 0.67V, A = I/2 = 0.33 V/r

So, V increases and A decreases.

Post the rest of the questions separately.

thanks.