Compare the power consumed by a 2 Ω resistor in each of the following circuits:

                                                     (OR)

Find the current and resistance drawn by a bulb rated 40 𝑊; 220 𝑉, when it is connected to a 220 𝑉 supply. If the given bulb is replaced by a bulb if rating 25 𝑊; 220 𝑉, will there be any change in the value of current and resistance? Justify your answer and determine the change.

Given resistance $\mathrm{V}=6\mathrm{V}, {\mathrm{R}}_1=1\mathrm{\Omega }\text{ and }{\mathrm{R}}_2=2\mathrm{\Omega }\text{ to find }\mathrm{l}\text{. }$

Total Resistance,

 $\begin{array}{r}{\mathrm{R}}_{\mathrm{s}}={\mathrm{R}}_1+{\mathrm{R}}_2\\ \Rightarrow {\mathrm{R}}_{\mathrm{s}}=1+2\\ \Rightarrow {\mathrm{R}}_{\mathrm{s}}=3\mathrm{\Omega }\end{array}$

According to Ohm's Law, V= IR_S

$\begin{array}{r}\Rightarrow \mathrm{I}=\frac{\mathrm{V}}{{\mathrm{R}}_{\mathrm{s}}}\\ \Rightarrow \mathrm{I}=\frac{6}{3}\\ \Rightarrow \mathrm{I}=2\mathrm{A}\end{array}$

In a series combination, the current is same for all resistors, but the potential difference differs. 

So, let the potential difference of 2 Ohm Resistor = V₁ volts. 

Let the power of 2 $\mathrm{\Omega }$ resistors = P₁W

$$\begin{gathered} {\mathrm{V}}_1={\mathrm{IR}}_2 \\ \Rightarrow {\mathrm{V}}_1=2\times 2 \\ \Rightarrow {\mathrm{V}}_1=4\mathrm{V} \end{gathered}$$

Electric Power = V X I

Power of 2 Ω Resistor = 𝑉₁ × I

$\begin{array}{r}{\mathrm{P}}_1=4\times 2\\ {\mathrm{P}}_1 =8\mathrm{W}\end{array}$

For case (ii): Let us consider total current in circuit = 𝐼 

In parallel combination V is same but I differ. 

Let total resistance = $R_p$

$\begin{array}{r}\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}\\ \Rightarrow \frac{1}{R_p}=\frac{1}{1}+\frac{1}{2}\\ \Rightarrow \frac{1}{R_p}=\frac{2+1}{2}\\ \Rightarrow \frac{1}{R_p}=\frac{3}{2}\\ \Rightarrow R_p=\frac{2}{3}\mathrm{\Omega }\end{array}$

According to Ohm's Law,

$\begin{array}{r}V=I{R}_p\\ \Rightarrow 4=I\times \frac{2}{3}\\ \Rightarrow 2=\frac{I}{3}\\ \Rightarrow I=6\mathrm{A}\\ \text{ Now, }{I}_2=\frac{V}{R_2}\\ \Rightarrow I_2=\frac{4}{2}\\ \Rightarrow I_2=2\mathrm{A}\end{array}$

$\mathrm{Let} \mathrm{the} \mathrm{power} \mathrm{of} {\mathrm{R}}_2={\mathrm{P}}_2\mathrm{W}$

$\begin{array}{r}\Rightarrow {\mathrm{P}}_2=\mathrm{V}\times {\mathrm{I}}_2\\ \Rightarrow {\mathrm{P}}_2=4\times 2\\ \Rightarrow {\mathrm{P}}_2=8\mathrm{W}\end{array}$

                                            (OR)

Case 1: power of bulb, P = 40 watt 

The voltage supplied across the bulb, 𝑉 = 220𝑉 

We know, power supplied by bulb, 𝑃 = 𝑉𝐼

or, 40 = 220 × 𝐼

$\Rightarrow$Current 𝐼 = 40/220 = 2/11 = 0. 18 𝐴𝑚𝑝 

Again, using the formula, 𝑃 = 𝑉²/𝑅 to find resistance of bulb

or, 𝑅 = 𝑉²/𝑃 

$\Rightarrow$𝑅 = (220)²/40 = 48400/40 = 1210 𝑜ℎ𝑚

Case 2: if the given bulb is replaced by a bulb operating 25 𝑊; 220 V

Current, 𝐼' = 𝑃'/𝑉 = 25/220 = 0. 1136 𝐴𝑚𝑝 

resistance, 𝑅' = 𝑉²/𝑃' = (220)²/25 = 1936 𝑜ℎ𝑚. 

When we change the bulb, current and resistance will be changed. 

Change in current = 𝐼 − 𝐼' = 0. 18 − 0. 1136 = 0. 0664 𝐴𝑚𝑝 

The current decreases when we use 25 watt bulb in place of 40 watt bulb. 

Change in resistance = 1936 − 1210 = 726 𝑜ℎ𝑚

 The resistance increase when we use 25 watt bulb in place of 40 watt.