Physics - Current Electricity Question with Solution | TestHub

The resistance of each bulb is $R=V_0^2/P_0$. A. In series, total resistance is $2R$. Power $P_S=V_0^2/(2R)=V_0^2/(2V_0^2/P_0)=P_0/2$. B. In parallel, equivalent resistance is $R/2$. Power $P_P=V_0^2/(R/2)=2V_0^2/R=2P_0$. C. In series with $V=2V_0$, total resistance $2R$. Power $P_S^{\prime }=(2V_0{)}^2/(2R)=4V_0^2/(2R)=2V_0^2/R=2P_0$. D. In parallel with $V=V_0/2$, equivalent resistance $R/2$. Power $P_P^{\prime }=(V_0/2{)}^2/(R/2)=(V_0^2/4)/(R/2)=V_0^2/(2R)=P_0/2$.