On bridges

For the Wheatstone bridge above, let R=39113Ω,R2=252Ω,R4=120Ω,VEX=5VR = \frac {391}{13}\:\Omega,\: R_2 = 252\:\Omega,\: R_4 = 120\:\Omega,\: V_{EX} = 5\: V dc. If VO=0V V_O = 0 \: V, and the voltage drop across RR is 19552743V \frac {1955}{2743} \: V , determine the sum (in ohms) of the resistances R1 R_1 and R3 R_3.

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