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.

# 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.

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### Solution:

Given resistance

Total Resistance,

According to Ohm's Law, V= IRS

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

Electric Power = V X I

Power of 2 Ω Resistor = 𝑉1 × I

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

In parallel combination V is same but I differ.

Let total resistance = ${R}_{p}$

$\begin{array}{r}\frac{1}{{R}_{p}}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}\\ ⇒\frac{1}{{R}_{p}}=\frac{1}{1}+\frac{1}{2}\\ ⇒\frac{1}{{R}_{p}}=\frac{2+1}{2}\\ ⇒\frac{1}{{R}_{p}}=\frac{3}{2}\\ ⇒{R}_{p}=\frac{2}{3}\mathrm{\Omega }\end{array}$

According to Ohm's Law,

(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

$⇒$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

$⇒$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.

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