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(C) The circuit consists of a $2 \, V$ battery connected to a delta network of three $30 \, \Omega$ resistors. One node of the triangle is connected t

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$\frac{1}{10} \, A$

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(C) The circuit consists of a $2 \, V$ battery connected to a delta network of three $30 \, \Omega$ resistors. One node of the triangle is connected to the negative terminal,and the opposite side is connected to the positive terminal. This results in two $30 \, \Omega$ resistors being in series,and this combination being in parallel with the third $30 \, \Omega$ resistor. Equivalent resistance $R_{eq} = \frac{(30 + 30) 	imes 30}{(30 + 30) + 30} = \frac{60 	imes 30}{90} = 20 \, \Omega$. Using Ohm's law,the current $i = \frac{V}{R_{eq}} = \frac{2}{20} = \frac{1}{10} \, A$.

The current $i$ in the adjoining circuit will be:

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