In the circuit shown, the resistances are given in ohms and the battery is assumed ideal with $\mathrm{emf}$ equal to $3.0$ $\mathrm{volts}.$ The resistor that dissipates the most power is
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Power is related to current and resistance by $P=I^{2} R \quad .$ The resistor that has
the most current would be $R_{1}$.
since all the other resistor share a current that is split from the main current running from battery to $R_{1}$
since $R_{e q} < R_{1},$ the most power is thus dissipated through $R_{1}$
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