Two tangent galvanometers A and B are identic | vedclass

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(A) The principle of a tangent galvanometer is given by $I = K 	an 	heta$,where $K = \frac{2rB_H}{\mu_0 N}$.Since the galvanometers are identical ex

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$1: 3$

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(A) The principle of a tangent galvanometer is given by $I = K 	an 	heta$,where $K = \frac{2rB_H}{\mu_0 N}$.Since the galvanometers are identical except for the number of turns $N$ and are connected in series,the current $I$ flowing through them is the same.Thus,$I = \frac{2rB_H}{\mu_0 N} 	an 	heta$,which implies $N 	an 	heta = 	ext{constant}$.Therefore,$N_A 	an 	heta_A = N_B 	an 	heta_B$.The ratio of the number of turns is $\frac{N_A}{N_B} = \frac{	an 	heta_B}{	an 	heta_A}$.Given $	heta_A = 60^{\circ}$ and $	heta_B = 30^{\circ}$,we have $\frac{N_A}{N_B} = \frac{	an 30^{\circ}}{	an 60^{\circ}}$.$\frac{N_A}{N_B} = \frac{1/\sqrt{3}}{\sqrt{3}} = \frac{1}{3}$.Thus,the ratio is $1: 3$.

Two tangent galvanometers $A$ and $B$ are identical except in their number of turns. They are connected in series. On passing a current through them,deflections of $60^{\circ}$ and $30^{\circ}$ are produced. The ratio of the number of turns in $A$ and $B$ is

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