Mr Bean,
The current in the circuit is 12V/10.8ohm=1.11A.
The voltage drop across the resistor is V=IR=1.11A*6.8ohm=7.48V
Using P=VI=7.48V*1.11A=8.3W
and voltage drop across motor is V=IR=1.11A*4.0ohm=4.44V
Using P=VI=4.44V*1.11A=4.9W
The motor & resistor see the same current but the voltage drop is different.
Make sense?
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