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The given network of resistances is a part of an electric circuit containing the sources of EMF providing current. The potential difference across $$6\ \Omega$$ resistance is measured to be 12 volt as shown in the figure. The value of potential difference across the terminal A and B is
The $$6\ \Omega$$ and the lower $$3\ \Omega$$ are in parallel, so both have 12 V, carrying 2 A and 4 A, i.e. a total current of 6 A through the network. Their combination is $$2\ \Omega$$, and the next block $$12 \parallel 12 \parallel 3 = 6 \parallel 3 = 2\ \Omega$$. The total resistance from A to B is $$4 + 2 + 2 + 2 = 10\ \Omega$$, so $$V_{AB} = 6 \times 10 = 60$$ V.
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