Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/410

 

$R$ rises from zero to its final value is expressed by a formula of exactly the same kind as that which expresses the value of a current urged by a constant electromotive force through the coil of an electromagnet. Hence we may place a condenser and an electromagnet on two opposite members of Wheatstone's Bridge in such a way that the current through the galvanometer is always zero, even at the instant of making or breaking the battery circuit.

In the figure, let $P$, $Q$, $R$, $S$ be the resistances of the four members of Wheatstone's Bridge respectively. Let a coil, whose coefficient of self-induction is $L$ be made part of the member $AH$, whose resistance is $Q$, and let the electrodes of a condenser, whose capacity is $C$, be connected by pieces of small resistance with the points $F$ and $Z$. For the sake of simplicity, we shall assume that there is no current in the galvanometer $G$, the electrodes of which are connected to $F$ and $H$. We have therefore to determine the condition that the potential at $F$ may be equal to that at $H$. It is only when we wish to estimate the degree of accuracy of the method that we require to calculate the current through the galvanometer when this condition is not fulfilled.

Let $x$ be the total quantity of electricity which has passed through the member $AF$, and $z$ that which has passed through $FZ$ at the time $t$, then $x-z$ will be the charge of the condenser. The electromotive force acting between the electrodes of the condenser is, by Ohm's law, $R{\frac {dz}{dt}}$, so that if the capacity of the condenser is $C$,

$x-z=RC{\frac {dz}{dt}}$.

(1)

Let $y$ be the total quantity of electricity which has passed through the member $AH$, the electromotive force from $A$ to $H$ must be equal to that from $A$ to $F$, or

$Q{\frac {dy}{dt}}+{\frac {d^{2}y}{dt^{2}}}=P{\frac {dx}{dt}}$.

(2)

Since there is no current through the galvanometer, the quantity which has passed through $HZ$ must be also $y$, and we find

$S{\frac {dy}{dt}}=R{\frac {dz}{dt}}$.

(3)

Substituting in (2) the value of $x$, derived from (1), and com paring with (3), we find as the condition of no current through the galvanometer

$RQ\left(1+{\frac {L}{Q}}{\frac {d}{dt}}\right)=SP\left(1+RC{\frac {d}{dt}}\right)$.

(4)