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

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