Three resistors $2 Ω, 3 Ω$ and $4 Ω$ are connected in series in a circuit. The total resistance of the circuit is

7 Ω

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8 Ω

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9 Ω

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10 Ω

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Solution

The correct option is
B

9 Ω Given: Resistances,
$R_1 = 2 ~Ω$ , $R_2 = 3 ~Ω$ and $R_3 = 4 ~Ω$ .

When all the resistors are in series, equivalent resistance of the circuit, $R_e = R_1 + R_2 + R_3 = 2 + 3 + 4 = 9 ~Ω$ .

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