Part (a)
For balance to occur there must be no current flowing in the Galvanometer, therefore, there no voltage drop across the #2.0Omega# resistor and the voltage is at the same potential as the negative terminal of #1.5" V"# battery. This means that the voltage across the x length, #R_x#, must be #-1.5" V"#
#V_(R_x) = -1.5" V"#
The current flowing through #R_x# is:
#I_(R_x) = (-2.0" V")/(1.0Omega+10Omega)#
Compute the value of #R_x#:
#R_x = V_(R_x)/I_(R_x)#
#R_x = (-1.5" V")/((-2.0" V")/(1.0Omega+10Omega))#
#R_x = 0.75(11Omega)#
#R_x = 8.25Omega#
Using the ratio, #X/R_x=(100" cm")/(10.0Omega)#:
#X = (100" cm")(8.25Omega)/(10.0Omega)#
#X = 82.5" cm"#
Part (b)
Compute the current supplied by the negative terminal of the #1.5" V"# battery with 0 current flowing through the Galvanometer:
#I_(1.5" V") = (-1.5" V")/(1Omega+2Omega+5Omega)#
#I_(1.5" V") = -0.1875" A"#
Compute the voltage at the Galvanometer:
#V_G = -1.5" V" - (-0.1875" A")1Omega - (-0.1875" A")2Omega#
#V_G = -0.9375" V"#
From part (a), we know the current that is flowing in #R_x#:
#I_(R_x) = (-2.0" V")/(1.0Omega+10Omega)#
We want #V_(R_x)# to be equal to #V_G#:
#V_(R_x) = -0.9375" V"#:
Compute the value of #R_x#
#R_x = (-0.9375" V")/((-2.0" V")/(1.0Omega+10Omega))#:
#R_x = 5.16Omega#
Using the ratio, #X/R_x=(100" cm")/(10.0Omega)#:
#X=(100" cm")(5.16Omega)/(10.0Omega)#:
#X = 51.6" cm"#
