The equivalent resistance of parallel resistors,

$\overline{)\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{eq}}}{\mathbf{=}}\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{1}}}{\mathbf{+}}\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{2}}}}$

Series resistors,

$\overline{){{\mathbf{R}}}_{\mathbf{e}\mathbf{q}}{\mathbf{=}}{{\mathbf{R}}}_{{\mathbf{1}}}{\mathbf{+}}{{\mathbf{R}}}_{{\mathbf{2}}}}$

$\overline{){\mathbf{V}}{\mathbf{=}}{\mathbf{I}}{\mathbf{R}}}$

The equivalent resistance of the bulbs,

${\mathit{R}}_{\mathbf{e}\mathbf{q}}\mathbf{=}{\mathbf{(}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{)}}^{\mathbf{-}\mathbf{1}}$

R_{eq} = **1.5 Ω**

R and R_{eq} are in series.

R_{R,Req} = **1.5 + R**

Two bulbs are connected in parallel across a source of emf ℰ = 11.0 V with a negligible internal resistance. One bulb has a resistance of 3.0 Ω , and the other is 3.0 Ω . A resistor R is connected in the circuit in series with the two bulbs. What value of R should be chosen in order to supply each bulb with a voltage of 2.4 V ?

For what value of *R* would the potential difference across each of the bulbs be 2.4 V ? Express your answer in ohms using three significant figures.

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