Question

A torch bulb is rated $2.5V$and $750mA$. Calculate its power, its resistance, and the energy consumed if this bulb is lighted for $4hours$. Two identical resistors, each of resistance $2\Omega$ are connected in the torch in series and in parallel, to a battery of $12V$, calculate the ratio of power consumed in two cases.

Answer 1

Step 1: Given

Torch Voltage = $2.5V$

Torch current = $750mA$

Time = $4hours$

Resistance = $2\Omega$

Voltage of battery = $12V$

Step 2: Formula used and calculation

Power is given by

$$\begin{gathered} P=VI \\ =2.5V\times .75A \\ =1.875W \end{gathered}$$

Power consumed by torch is $1.875W$.

Resistance is given by

$$\begin{gathered} R=\frac{V}{I} \\ =\frac{2.5V}{0.75A} \\ =3.33\Omega \end{gathered}$$

Resistance of torch is $3.33\Omega$.

Energy consumed is given by

$$\begin{gathered} E=Pt \\ =1.875W\times 4\times 60\times 60s \\ =27000J \end{gathered}$$

In series combination

$$\begin{gathered} R_{eq}=R_1+R_2 \\ =2\Omega +2\Omega \\ =4\Omega \end{gathered}$$

$$\begin{gathered} V=12V \\ I=\frac{V}{R_{eq}} \\ =\frac{12}{4}A \\ =3A \\ {P}_1=VI \\ =12\times 3W \\ =36W \end{gathered}$$

In parallel combination

$$\begin{gathered} \frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2} \\ =\frac{1}{2}+\frac{1}{2}\Omega \\ {R}_{eq}=1\Omega \\ V=12V \\ I=\frac{V}{R_{eq}} \\ =\frac{12}{1}A \\ =12A \\ {P}_2=VI \\ =12\times 12W \\ =144W \end{gathered}$$

Ratio of power in series and parallel combination

$$\begin{gathered} \frac{P_1}{P_2}=\frac{36}{144} \\ =\frac{1}{4} \end{gathered}$$

Hence, the ratio is $1:4$.