In the circuit as shown in the figure, the:

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(D) First, calculate the equivalent resistance $R'$ of the three parallel resistors $(10\,\Omega, 10\,\Omega, 20\,\Omega)$:$\frac{1}{R'} = \frac{1}{10

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All options are correct

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(D) First, calculate the equivalent resistance $R'$ of the three parallel resistors $(10\,\Omega, 10\,\Omega, 20\,\Omega)$:$\frac{1}{R'} = \frac{1}{10} + \frac{1}{10} + \frac{1}{20} = \frac{2+2+1}{20} = \frac{5}{20} = \frac{1}{4} \Rightarrow R' = 4\,\Omega$.Using Ohm's law for the total circuit, where the total current $I = 0.5\,A$ and total voltage $V = 25\,V$:$I = \frac{V}{R + R'} \Rightarrow 0.5 = \frac{25}{R + 4}$.$R + 4 = \frac{25}{0.5} = 50 \Rightarrow R = 46\,\Omega$.Now, calculate the potential difference across the parallel combination:$V_{parallel} = I 	imes R' = 0.5 	imes 4 = 2\,V$.Since the resistors are in parallel, the potential difference across each is $2\,V$.Check the current through the $20\,\Omega$ resistor:$I_{20} = \frac{V_{parallel}}{20} = \frac{2}{20} = 0.1\,A$.Since all statements are correct, the correct option is $D$.

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