Electrical Engineering All Engineering Career
Question 61
Switching of a lamp in house produces noise in the radio. This is because switching operation produces ?
The switching of a lamp in a house producing noise in the radio is because the switching operation produces arcs across separating contacts. These arcs can generate electromagnetic interference (EMI), which can be picked up by nearby electronic devices such as radios.
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Question 62
Which method can be used for absolute measurement of resistances ?
The method that can be used for the absolute measurement of resistances is the Wheatstone bridge method. This method allows for precise measurement of resistance by comparing the unknown resistance with known resistances in a balanced bridge circuit.
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Question 63
Which of the following can have negative temperature coefficient ?
Among the options provided, electrolytes can have a negative temperature coefficient. In certain electrolytes, the conductivity decreases with an increase in temperature, leading to a negative temperature coefficient of resistance.
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Question 64
Three identical resistors are first connected in parallel and then in series. The resultant resistance of the first combination to the second will be ?
When three identical resistors are connected in parallel, the equivalent resistance (\(R_{\text{parallel}}\)) is given by:
\[ \dfrac{1}{R_{\text{parallel}}} = \dfrac{1}{R} + \dfrac{1}{R} + \dfrac{1}{R} \]
\[ \dfrac{1}{R_{\text{parallel}}} = \dfrac{3}{R} \]
\[ R_{\text{parallel}} = \dfrac{R}{3} \]
When three identical resistors are connected in series, the equivalent resistance (\(R_{\text{series}}\)) is given by:
\[ R_{\text{series}} = R + R + R = 3R \]
Now, to find the ratio \( \dfrac{R_{\text{parallel}}}{R_{\text{series}}} \):
\[ \dfrac{R_{\text{parallel}}}{R_{\text{series}}} = \dfrac{\dfrac{R}{3}}{3R} \]
\[ \dfrac{R_{\text{parallel}}}{R_{\text{series}}} = \dfrac{1}{9} \]
So, the resultant resistance of the first combination to the second is \( \dfrac{1}{9} \) times.
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Question 65
Resistance of 220 V, 100 W lamp will be ?
To find the resistance of the lamp, we can use the formula:
\[ P = \dfrac{V^2}{R} \]
Where:
- \( P \) is the power (in watts)
- \( V \) is the voltage (in volts)
- \( R \) is the resistance (in ohms)
Given:
- \( P = 100 \) W
- \( V = 220 \) V
We rearrange the formula to solve for \( R \):
\[ R = \dfrac{V^2}{P} \]
\[ R = \dfrac{220^2}{100} \]
\[ R = \dfrac{48400}{100} \]
\[ R = 484 \, \Omega \]
So, the resistance of the lamp is 484 ohms.
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Question 66
With three resistances connected in I parallel, if each dissipates ,20 W the total power supplied by the voltage source equals ?
When three resistances are connected in parallel, the total power dissipated is the sum of the power dissipated by each resistor. Since each resistor dissipates 20 W and there are three resistors, the total power dissipated is:
\[ 20 \, \text{W} + 20 \, \text{W} + 20 \, \text{W} = 60 \, \text{W} \]
So, the total power supplied by the voltage source is 60W..
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Question 67
which resistor will be physically larger in size ?
The physical size of a resistor depends on both its resistance value and its power rating. Generally, higher power ratings and higher resistance values require larger physical sizes to dissipate heat effectively.
Comparing the given options:
1. \(100 \, \Omega\), \(10 \, \text{W}\)
2. \(10 \, \Omega\), \(50 \, \text{W}\)
3. \(1 \, \text{M}\Omega\), \(1/2 \, \text{W}\)
4. \(1 \, \text{k}\Omega\), \(1 \, \text{W}\)
Among these options, the resistor with the highest power rating is the \(10 \, \Omega\), \(50 \, \text{W}\) resistor. Therefore, it is likely to be physically larger in size compared to the others.
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Question 68
A resistance having rating 10 ohms, 10 W is likely to be a ?
A resistance with a rating of 10 ohms and 10 W is likely to be a wire-wound resistor. Wire-wound resistors are commonly used for high-power applications due to their ability to handle large amounts of power without overheating.
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Question 69
Which of the following lamps will have least resistance at room temperature ?
The resistance of a lamp can be calculated using Ohm's law:
\[ R = \dfrac{V^2}{P} \]
Where:
- \( R \) is the resistance (in ohms)
- \( V \) is the voltage (in volts)
- \( P \) is the power (in watts)
Let's calculate the resistance for each lamp:
1. For the 26 W lamp:\[ R = \dfrac{220^2}{26} \approx 1865.38 \, \Omega \]
2. For the 100 W lamp:\[ R = \dfrac{220^2}{100} = 484 \, \Omega \]
3. For the 200 W lamp:\[ R = \dfrac{220^2}{200} = 242 \, \Omega \]
4. For the 60 W lamp:\[ R = \dfrac{220^2}{60} \approx 805.67 \, \Omega \]
Among these options, the lamp with the least resistance at room temperature is the 200 W, 220 V lamp, with a resistance of 242 ohms.
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Question 70
Copper wire of certain length and resistance is drawn out to three times its length without change in volume, the new resistance of wire becomes ?
When a copper wire is drawn out to three times its length without a change in volume, its cross-sectional area decreases by a factor of 9 (since volume remains constant).
The resistance of a wire is inversely proportional to its cross-sectional area. Therefore, if the cross-sectional area decreases by a factor of 9, the resistance increases by a factor of 9.
So, the new resistance of the wire becomes 9 times its original resistance.