When a resistor with resistance R is connected to a 1.50-V flashlight

Step 1
Given
We are given the emf of the battery ${\displaystyle \xi =1.50V}$, where the resistor consumes power ${\displaystyle P=0.0625W}$. Also, we can neglect the internal resistance of all batteries where ${\displaystyle r=0}$.
Solution
(a) We want to find the power consumed by the resistor if it is connected to a battery with emf ${\displaystyle \xi =12.6V}$. To get the target we should find the resistance of the resistor where the resistance is constant as we are given.
The power consumed by the resistor in the circuit depends on the resistance R of the resistor and it is given by equation 25.18 in the form ${\displaystyle P=\frac{V^2}{R}}$ (1)
Where V is the terminal voltage in the circuit and in our case the internal resistance could be neglible, so the terminal voltage equals the emf of the battery ${\displaystyle V=\xi =1.50V}$.
Now let us solve equation (1) to R and plug our values for V and P into equation (1) to get R
${\displaystyle R=\frac{V^2}{P}=\frac{{\left(1.50V\right)}^2}{0.0625W}=36\mathrm{\Omega }}$
The resistance is the same, so if the battery voltage is 12.6V, therefore we can use equation (1) to get the consumed power by the resistor as next
${\displaystyle P=\frac{V^2}{R}=\frac{{\left(12.6V\right)}^2}{36\mathrm{\Omega }}=4.42W}$
Step 2
(b) We want to find the emf if the power consumed is ${\displaystyle P=5W}$. By the same concept, if we neglect the internal resistance of the battery the emf equals the terminal voltage. So we can use equation (1) to get voltage (emf) with the same value of resistance
${\displaystyle R=\frac{V^2}{P}}$
${\displaystyle V=\sqrt{PR}}$
${\displaystyle V=\sqrt{\left(5W\right)\left(36\mathrm{\Omega }\right)}}$
${\displaystyle V=13.41V}$
The emf ${\displaystyle \xi =13.41V}$

The key to unlock this complicated mystery is to know the value of the resistor. We can calculate it from the description of Scene 1:
Power dissipated by a resistance $=(voltage{)}^2/(resistance)$
Resistance $=(voltage{)}^2/(power)$
In scene 1:
Resistance ${\displaystyle =\frac{{\left(1.5v\right)}^2}{0.0625w}}$
Resistance ${\displaystyle =36Ohms}$
That helps a bunch ! It's all we need to solve the second part.
$Power=(voltage{)}^2/(resistance)$
${\displaystyle Power=\frac{{\left(12.6v\right)}^2}{36\mathrm{\Omega }}}$
Power ${\displaystyle =4.41wa\mathtt{s}}$
Another (easier) way:
$Power=(voltage{)}^2/(resistance)$
SO ... if the resistance doesn't change, then the power is proportional to the square of the voltage.
$(powe{r}_2/powe{r}_1)=(voltag{e}_2/voltag{e}_1{)}^2$
$powe{r}_2=(powe{r}_1)\cdot (voltag{e}_2/voltag{e}_1{)}^2$
${\displaystyle powe{r}_2=\left(0.0625w\right)\cdot {\left(12.6\frac{v}{1.5}v\right)}^2}$
${\displaystyle powe{r}_2=\left(0.0625w\right)\cdot {\left(8.4\right)}^2}$
${\displaystyle powe{r}_2=\left(0.0625w\right)\cdot \left(70.56\right)}$
${\displaystyle powe{r}_2=4.41w}$

(a) Given info: First the reistance (R) is connected across 1.50 V battery voltage $(V_1)$, consumes power $(P_1)$ of 0.0625W. And next the resistance (R) connected across 12.6V battery voltage $(V_2)$, it consumes power $(P_2)$.
The formula for the power $(P_1)$ consumed by the resistor (R) when it is connected across 1.50V battery voltage $(V_1)$ is,
$P_1=\frac{V_1^2}{R}$(I)
- $(V_1)$ is first battery voltage,
- R is the resistance of the resistor.
Rearrange the equation (I) to find R.
$R=\frac{V_1^2}{P_1}$ (II)
The formula for the power $(P_2)$ consumed by the resistor (R) when it is connected across 12.6V battery voltage $(V_2)$ is,
$P_2=\frac{V_2^2}{R}$ (III)
- $(V_2)$ is the second battery voltage
NSK
Rearrange the equation (I)
$R=\frac{V_2^2}{P_2}$ (IV)
Since reistance (R) remains constant for both the cases, equate equation (II) and (IV).
$\frac{V_1^2}{P_1}=\frac{V_2^2}{P_2}$
Rearrange the equation (V) to find power $(P_2)$,
$P_2=P_1(\frac{V_2^2}{V_1^2})$ (VI)
$=P_1(\frac{V_2}{V_1}{)}^2$
Substitute 1.50V for $(V_1)$, 12.6V for $(V_2)$ and 0.0625W for $(P_1)$ to find $(P_2)$,
$$\begin{gathered} P_2=(0.0625W)(\frac{12.6V}{1.50V}{)}^2 \\ =(0.0625W)(70.56) \\ =4.41W \end{gathered}$$
The power consumption $(P_2)$ of resistance (R) when it is connected across 12.6 V battery voltage $(V_2)$ is 4.41W.
(b) Given info: The resistor is connected across battery voltage of $(V_2)$ with the power consumption $(P_2)$ of 5.00W.
Rearrange the equation (V).
$V_2^2=V_1^2(\frac{P_2}{P_1})$
Take square root on both the sides.
$V_2=V_1\sqrt{\frac{P_2}{P_1}}$
Substitute 1.50V for $(V_1)$, 0.0625W for $(P_1)$ and 5.00W for $(P_2)$ and to find $(P_2)$,
$$\begin{gathered} V_2=(1.50V)\sqrt{\frac{5.00W}{0.0625W}} \\ =(1.50V)(\sqrt{80}) \\ =(1.50V)(8.94) \\ =13.4V \end{gathered}$$
The voltage $(V_2)$ of resistance (R) when it consumes power $(P_2)$ of 5.00W is 13.4.
Conclusion:
Therefore, the voltage $(V_2)$ of resistance (R) when it consumes power $(P_2)$ of 5.00W is 13.4V.