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(B) The total resistance $R_{eq}$ of the circuit is given by Ohm's law: $R_{eq} = \frac{V}{I} = \frac{100 \,V}{2.5 \,A} = 40 \,\Omega$.Each resistor h

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$5, 5$

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(B) The total resistance $R_{eq}$ of the circuit is given by Ohm's law: $R_{eq} = \frac{V}{I} = \frac{100 \,V}{2.5 \,A} = 40 \,\Omega$.Each resistor has a resistance $R = 100 \,\Omega$. The equivalent resistance of $n_1$ resistors in parallel is $R_1 = \frac{R}{n_1} = \frac{100}{n_1}$.The equivalent resistance of $n_2$ resistors in parallel is $R_2 = \frac{R}{n_2} = \frac{100}{n_2}$.Since the two groups are in series,$R_{eq} = R_1 + R_2 = \frac{100}{n_1} + \frac{100}{n_2} = 40$.Dividing by $20$,we get $\frac{5}{n_1} + \frac{5}{n_2} = 2$.Given that $n_1 + n_2 = 10$,we test the options:For option $B$,$n_1 = 5$ and $n_2 = 5$: $\frac{5}{5} + \frac{5}{5} = 1 + 1 = 2$. This satisfies the equation.Thus,the correct values are $n_1 = 5$ and $n_2 = 5$.

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