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Basic questions on the Wheatstone Bridge

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The following circuit shows a Wheatstone Bridge with an unknown resistance. 

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a) The unknown resistance in a wheatstone bridge is equal to: $R_x = \\frac{R_1\\,R_4}{R_2} = \\frac{3\\,\\text{k}\\Omega\\times2\\,\\text{k}\\Omega}{5\\,\\text{k}\\Omega} = 1.2\\,\\text{k}\\Omega$. Where $R_1 = 3\\,\\text{k}\\Omega,\\;R_2=5\\,\\text{k}\\Omega,\\;R_4=2\\,\\text{k}\\Omega$.

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b) Rearranging the same formula, \\[R_2 \\times R_3 = R_1 \\times R_x\\\\R_x = \\frac{5\\,\\text{k}\\Omega\\times 2\\,\\text{k}\\Omega}{3\\,\\text{k}\\Omega} = 3.33\\,\\text{k}\\Omega\\]

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c) When the Wheatstone bridge is balanced, $I_A = 0$.

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d) The ratio of the resistances determine whether the bridge remains balanced, the source voltage does not affect the bridge. Therefore, the bridge remains balanced. 

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Given that the bridge is balanced, what is the value of the unknown resistance Rx?

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[[0]]k$\\Omega$

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What should the value of the variable resistance Rx be for the bridge to be balanced?

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[[0]] k$\\Omega$

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When the bridge is balanced, the value of IA is [[0]].

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If the batter is replaced with another battery which can provide 4 times the e.m.f. will the bridge remain balanced?

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