A circular coil of total resistance 40 Omega | vedclass

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(D) The circular coil is divided into two parts by points $P$ and $Q$ with resistances $R_1 = 30 \Omega$ and $R_2 = 10 \Omega$. These two parts are co

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$2$

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(D) The circular coil is divided into two parts by points $P$ and $Q$ with resistances $R_1 = 30 \Omega$ and $R_2 = 10 \Omega$. These two parts are connected in parallel between points $P$ and $Q$.The equivalent resistance $R_{PQ}$ of the parallel combination is given by:$\frac{1}{R_{PQ}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{30} + \frac{1}{10} = \frac{1+3}{30} = \frac{4}{30}$$R_{PQ} = \frac{30}{4} = 7.5 \Omega$The total resistance of the circuit including the internal resistance $r = 0.5 \Omega$ is:$R_{total} = R_{PQ} + r = 7.5 \Omega + 0.5 \Omega = 8 \Omega$The current $I$ flowing through the circuit is given by Ohm's law:$I = \frac{V}{R_{total}} = \frac{16 	ext{ V}}{8 \Omega} = 2 	ext{ A}$

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