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Physics Topics cover a broad range of concepts that are essential to understanding the natural world.
What do you mean by watt rating of a resistor? What do you mean by the statement ‘rating of a resistor is 1W-100Ω?
Voltage rating: Each and every electrical instrument depends on potential difference. Each instrument is marked with a fixed value of potential difference, indicating the range of potential difference within which it can be run safely. This is called voltage rating.
For example, the voltage rating of each domestic appliance is 220 V. This means that, if the potential difference between two ends of the appliance is 220 V, then its efficiency becomes max-imum. The efficiency of the appliance decreases if for any reason the supply voltage is diminished, i.e., the electric lamp glows dim, the electric fan moves comparatively slowly etc. On the other hand, if the voltage exceeds 220 V, there is a chance of damage to the appliance.
This is to note that, the appliances having same voltage rating should be connected in parallel while using simultaneously in a circuit. In that case, the potential difference across each appli-ance remains same. For this reason, daily appliances are all con-nected in parallel combination with 220 V supply line.
Watt rating: Different electrical appliances, even running at the same voltage may not consume same electrical energy in unit time. Hence, the power of the appliance, i.e., electrical energy consumed per second should be mentioned. This is called watt rating of electrical appliances. Thus 220V-100 W on the body of a lamp or 220V-2000W on the body of a heater, refers to the voltage and watt ratings of the appliance.
Significance of ratings:
i) Maximum capability: Voltage rating indicates at what voltage, the appliance does maximum work and watt rating indicates the rate of consumption of electrical energy at that voltage. For example, 220 V-100 W indicates that, if the voltage at the two ends of the lamp is 220 V, it will glow with maximum brightness, consuming energy at the rate of 100 W.
ii) Calculation of current and resistance: It is possible to cal-culate current and resistance of the appliance, when it operates with maximum efficiency.
Example:
I(current) \(=\frac{P(\text { power })}{V(\text { potential difference })}\) = \(\frac{100 \mathrm{~W}}{220 \mathrm{~V}}\) = 0.45 A
R (resistance) = \(\frac{V}{I}\) = \(\frac{220 \mathrm{~V}}{0.45 \mathrm{~A}}\) = 489 Ω
iii) Safety of supply line: Whether an electrical appliance is safe or not for a supply line, is known from its rating. Consider the case of a 220 V-2000 W electric heater.
Current drawn by the heater = \(\frac{2000}{220}\) = 9.1 A (approx.)
So, this heater cannot be used In the house, where the meter and fuse are installed for 5 A current. If it is used, the supply line will be damaged.
Power in a combination of more than one electrical devices:
Parallel combination: Suppose, R1 and R2 are the resistances of two electrical appliances. Their watt ratings are P1 and P2 respectively and voltage rating of both of them is V. If they are connected in parallel to an electric source of voltage V [Fig. (a)], then the voltage at the two ends of both the appliances will be V. The two appliances will function at full power and the total power consumed by the circuit will be,
P = P1 + P2
For any number of electrical appliances connected in parallel, P = P1 + P2 + P3 + ….. obviously, total power will be greater than that of each appliance.
Household electrical appliances are usually connected each in parallel.
Now, P1 = \(\frac{V^2}{R_1}\) and P2 = \(\frac{V^2}{R_2}\)
∴ \(\frac{P_1}{P_2}\) = \(\frac{R_2}{R_1}\) …. (1)
For example, if two electric lamps of 220 v – 100 W and 220 V – 60 W are connected in parallel and the combination is joined to a 220 V line, the 100 W lamp will glow brighter and
the total power of the circuit will be, P = 100 + 60 = 160W.
Series combination: Now the two appLiances are connected in series and this combination is connected to the same electric source (Fig.(b)). Same current, say I, will flow through both of them. So, their powers will be respectively,
\(P_1^{\prime}\) = I2R1 and \(P_2^{\prime}\) = I2R2
So, \(\frac{P_1^{\prime}}{P_2^{\prime}}\) = \(\frac{R_1}{R_2}\) ….. (2)
From equations (1) and (2) we have,
\(\frac{p_1^{\prime}}{P_2^{\prime}}\) = \(\frac{P_2}{P_1}\)
So, if P1 > P2, \(P_1^{\prime}\) < \(P_2^{\prime}\)
i.e., in series combination, the appliance having higher watt rating will consume less power. For example, if two electric lamps having power 100 W and 60 W are connected ¡n series, the power of 1oo W lamp will be comparatively lower and it will glow with less brightness.
Again, according to the Fig.(b),
I = \(\frac{V}{R_1+R_2}\)
So, total power of the circuit,
For any number of electrical appliances connected in series,
\(\frac{1}{P^{\prime}}\) = \(\frac{1}{P_1}\) + \(\frac{1}{P_2}\) + \(\frac{1}{P_3}\) + ……
Obviously, total power P’ will be less than that of each appliance.
For example, if two electric lamps of 220 V – 100 W and 220 V – 60 W are connected in series and joined to a 220 V supply line, the 60 W lamp will glow brighter and the total power of the circuit will be less than 60 W. Here,
\(\frac{1}{P^{\prime}}\) = \(\frac{1}{100}+\frac{1}{60}\) or, p’ = \(\frac{100 \times 60}{100+60}\) = 37.5 W
Rating of a resistor: No resistor can endure excess current over a certain designated value depending upon its nature. If the limit is exceeded, the excessive heat generated in the resis
tor will burn it. We may say that, as soon as the power consumption due to production of heat crosses a certain limit, the resistor gets damaged. This maximum power tolerance is known as watt rating of the resistor.
For example, a resistor with rating 1W – 100Ω means a 100Ω resistance and of maximum power tolerance 1W. We know, power, P = VI = IR ᐧ I = I2R
So maximum current tolerance of the resistor is,
I = \(\sqrt{\frac{P}{R}}\) = \(\sqrt{\frac{1}{100}}\) = \(\frac{1}{10} \mathrm{~A}\)
Under this condition, potential difference between the two ends of the resistor is,
V = IR = \(\frac{1}{10}\) × 100 = 10V
Therefore, before joining in a circuit, the rating of a resistor should be known. In that case, it can be connected in such a way that current should always remain below \(\frac{1}{10}\)A through it.
Numerical Examples
Example 1.
A 220V-60W electric bulb is connected in 220 V line. What is the resistance of the filament of the bulb, when it is turned on?
Solution:
Power, P = VI = V ᐧ \(\frac{V}{R}\) = \(\frac{V^2}{R}\) [∵ V = IR]
∴ R = \(\frac{V^2}{P}\) = \(\frac{220 \times 220}{60}\) = 806.67Ω
Example 2.
The resistance of a hot tungsten filament is about 10 times that in its normal state. What will be the resistance of a 100 W – 200 V tungsten lamp in its normal state? (AIEEE ‘05)
Solution:
Resistance of the lamp when it is hot,
R = \(\frac{V^2}{P}\) = \(\frac{(200)^2}{100}\) = 400Ω
So, resistance of the lamp in its normal state = \(\frac{400}{10}\) = 4Ω.
Example 3.
The main meter of a house is marked 10 A – 220 V. How many 60 W electric lamps can be used safely In this line?
Solution:
Maximum power of the household line,
P = VI = 220 × 10 = 2200 W
Maximum number of 60 W lamp = \(\frac{2200}{60}\) = 36.67
So, maximum 36 lamps can be used at a time.
Example 4.
A 220 V-100 W electric lamp fuses above 150 W power. What should be the maximum tolerable voltage for the lamp? [HS ‘07]
Solution:
Power, P = \(\frac{V^2}{R}\)
If we ignore the change of resistance with the increase of power, we have, P ∝ V2
So, for two different powers, \(\frac{P_1}{P_2}\) = \(\frac{V_1^2}{V_2^2}\)
i.e., V2 = V1\(\sqrt{\frac{P_2}{P_1}}\) = 220 × \(\sqrt{\frac{150}{100}}\) = 269.4 V
So, the maximum tolerable voltage for the lamp is 269.4 V.
Example 5.
In order to run a 60V-120W lamp safely in a 220V dc line, a resistor of what minimum magnitude should be placed in series with it?
Solution:
When the lamp runs safely in the given dc line, the maximum current is,
I = \(\frac{120}{60}\) = 2A
So, this 2 A current will also flow through the resistance placed in series.
Again, potential difference at the two ends of the resistance
= 220 – 60 = 160 V.
The minimum value of the resistance = \(\frac{160}{2}\) = 80 Ω.
Example 6.
Draw a household circuit having a 1200 W toaster, a 1000W oven,a800W heater and a 1500W cooler. The circuit has a heavy duty wire and a 20 A circuit breaker. Will the circuit breaker trip, if all the appliances are operated simultaneously in a 220 V supply voltage?
Solution:
The household circuit is shown in the Fig. When all the appliances are operated simultaneously, the total power is, P = 1200 + 1000 + 800 + 1500 = 4500 W.
But the circuit breaker is rated for 20 A. So, if all the appliances are operated simultaneously, the circuit breaker will trip.
Example 7.
Two lamps of 200 W and 100 W are connected in series in 200 V mains. Assuming the resistance of the two lamps to remain unchanged, calculate the power consumed by each of them.
Solution:
P = \(\frac{V^2}{R}\) or, R = \(\frac{V^2}{P}\)
So, resistance of the first lamp, R1 = \(\frac{200 \times 200}{200}\) = 200 Ω
and resistance of the second lamp, R2 = \(\frac{200 \times 20}{100}\) = 400 Ω
Now, if we connect the two lamps in series with 200 V mains, current flowing through each of them,
I = \(\frac{V}{R_1+R_2}\) = \(\frac{200}{200+400}\) = \(\frac{1}{3} \mathrm{~A}\)
∴ Power of the first lamp, P1 = I2R1 = \(\frac{1}{3}\) × \(\frac{1}{3}\) × 200 = 22.2 W
Power of the second lamp, P2 = I2R2 = \(\frac{1}{3}\) × \(\frac{1}{3}\) × 400 = 44.4 W
Example 8.
Two electric bulbs each designed to operate with a power of 500 W in a 220 V line are connected in series in a 110V line. What will be the power generated by each bulb? [HS ‘02]
Solution:
Power, P = \(\frac{V^2}{R}\)
So, resistance of each bulb,
R = \(\frac{V^2}{P}\) = \(\frac{(220)^2}{500}\) = \(\frac{484}{5} \Omega\)
When connected iñ series, potential difference at the two ends of each bulb = \(\frac{110}{2}\) = 55V.
∴ Power of each bulb = \(\frac{(55)^2}{\frac{484}{5}}\) = \(\frac{55 \times 55 \times 5}{484}\) = 31.25 W
Example 9.
If the supply voltage drops from 220 V to 200 V, what would be percentage reduction in heat produced by a 220 V – 1000W heater? Neglect the change of resistance. If the change of resistance is taken into consideration, would the reduction of heat produced be smaller or larger than the previously calculated value? Explain.
Solution:
We know, P = \(\frac{V^2}{R}\)
So, R = \(\frac{V^2}{P}\)
If the variation of resistance is ignored,
in the first case, R = \(\frac{V_1^2}{P_1}\);
in the second case, R = \(\frac{V_2^2}{P_2}\)
So, the power of the heater will reduce by 17.36%; i.e., the percentage reduction of heat produced is 17.36%. In the above calculation, decrease of resistance with voltage drop was ignored. But actually, with decrease in temperature of the coil, its resistance also decreases. Hence according to the relation P = \(\frac{V^2}{R}\), power increases, i.e., heat supplied by the heater also increases. So, the reduction of heat produced will be lower than 17.36%.
Example 10.
In Fig., the emf of the cell, E = 20V. Rating of each resistance R1 and R2 is 1W – 100Ω. What should be the minimum value of the resistance R in the circuit? Also find Its minimum watt rating.
Solution:
According to the formula,
P = I2R
or, I = \(\sqrt{P / R}\)
The maximum current that can flow through each of the resistance R1 and R2 is,
I1 = I2 = \(\sqrt{\frac{1}{100}}\) = \(\frac{1}{10} \mathrm{~A}\)
∴ Maximum current that can flow through the resistance R,
I = I1 + I2 = 2 × \(\frac{1}{10}\) = \(\frac{1}{5}\)A
On the other hand, equivalent resistance of R1 and R2,
i.e., to keep the value of I lower than \(\frac{1}{5}\)A, R should be greater than 50Ω. Power consumption for the minimum value of R,
P = I2R = \(\left(\frac{1}{5}\right)^2\) × 50 = 2W
So, minimum value of R is 50Ω and its minimum watt rating is 2W.
Example 11.
If a 220V – 1000W lamp is connected in 110V line then what will be the power consumed by it?
Solution:
Resistance of the lamp, R = \(\frac{V_1^2}{P}\) = \(\frac{(220)^2}{1000} \Omega\)
∴ Power consumed by the lamp when it is connected in 110V line,
P2 = \(\frac{V_2^2}{R}\) = \(\frac{(110)^2 \times 1000}{(220)^2}\) = 250 W
Example 12.
2.2 kW power is supplied through a line of 10Ω resistance under 22000 V voltage difference. What is the rate of heat dissipation in the line?
Solution:
I = \(\frac{P}{V}\) = \(\frac{2.2 \times 10^3}{22000}\) [∵ 2.2 kW = 2.2 × 103 W]
= 0.1 A
∴ Rate of heat dissipation = I2R = (0.1)2 × 1o = o.1 W
Example 13.
The potential difference between the two ends of an electric lamp is decreased by 1 %. Neglecting the change in its resistance, calculate the percentage increase or decrease In the power of the lamp.
Solution:
We know power, P = \(\frac{V^2}{R}\)
∴ In P = 2 ln V – ln R
By differentiating, \(\frac{d P}{P}\) = 2 ᐧ \(\frac{d V}{V}\) [∵ R = constant, dR = 0]
According to the problem, \(\frac{d V}{V}\) = -1% = –\(\frac{1}{100}\)
∴ \(\frac{d P}{P}\) = –\(\frac{2}{100}\) = -2%
i.e., the power of the lamp will be decreased by 2%.
Example 14.
15 kW power is supplied through a line of 0.5Ω resistance under 250 V potential difference. Find the efficiency of the supply in percentage.
Solution:
Power, P = 15kW = 15000W
So, current in the supply line,
I = \(\frac{P}{V}\) = \(\frac{15000}{250}\) = 60A
Power loss due to inactive resistance in the line
= I2R = (60)2 × 0.5 = 1800W;
Therefore, total power in line 15000 + 1800 = 16800 W
∴ Efficiency \(=\frac{\text { effective power }}{\text { total power }}\) × 100%
= \(\frac{15000}{16800}\) × 100% = 89%
Example 15.
Two incandescent lamps (25 W, 120 V) and (100 W, 120 V) are connected in serles across a 240 V supply. Assuming that the resistances of the iamps do not vary with current, find the power dissipated in each lamp after the connection. [HS ‘11]
Solution:
The resistance of the lamp 25 W – 120 V,
R1 = \(\frac{120^2}{25}\) = 576Ω
The resistance of the lamp 100 W – 120 V, R2 = \(\frac{120^2}{100}\) = 144Ω
When the lamp is connected in series, the equivalent resistance of the circuit,
R’ = 576 + 144 = 720Ω
∴ Current flowing through the circuit = \(\frac{240}{720}\) = 0.33 A
∴ Power dissipated in the first bulb = (0.33)2 × 576 = 62.72 W
∴ Power dissipated in the second bulb = (0.33)2 × 144 = 15.68 W