A projector lamp can be used at a maximum voltage of 60 V, its resistance is 20, the series resistance (in ohms) required to operate the lamp from a 75 V supply is
A. 2
B. 3
C. 4
D. 5
Answer
598.2k+ views
Hint: To solve this given problem, use Ohm’s law which gives the relationship between voltage, current and resistance. Now, substitute the values in this formula for maximum voltage and obtain the value of current flowing through the lamp. Then, substitute this obtained current and operating voltage in the mentioned and find the resistance, Now, subtract both the value of resistances and find the additional resistance required.
Formula used:
$V=IR$
Complete answer:
Given: Maximum Voltage, ${V}_{max}= 60 V$
For maximum voltage, resistance required, ${R}_{1}= 20\Omega$
Operating voltage, ${V}_{operate}= 75 V$
We know, according to Ohm’s law,
$V=IR$ …(1)
Where, V is the potential difference
I is the current flowing
R is the resistance
For maximum voltage, the equation. (1) can be written as,
${V}_{max}=I{R}_{1}$
Substituting values in above equation we get,
$60= I \times 20$
$\Rightarrow I= 3 A$
Now, to operate the lamp from 75 V supply, equation. (1) can be written as,
${V}_{operate}= I{R}_{2}$
$\Rightarrow 75= 3{R}_{2}$
$\Rightarrow {R}_{2}= 15 \Omega$
Therefore, the additional resistance required will be,
$R= {R}_{2}-{R}_{1}$
Substituting values in above equation we get,
$R= 20-15$
$\Rightarrow R=5\Omega$
Thus, the series resistance (in ohms) required to operate the lamp from a 75 V supply is $5\Omega$.
So, the correct answer is option D i.e. 5.
Note:
Students must remember that there are few limitations of Ohm’s law. Ohm’s law is not applicable for unilateral electrical elements like diodes and transistors as they allow the flow of current in only one direction. It is also not necessary that all conductors obey Ohm’s law. There are semiconductors such as silicon and germanium which do not obey Ohm’s law and these semiconductors are known as Non-ohmic conductors.
Formula used:
$V=IR$
Complete answer:
Given: Maximum Voltage, ${V}_{max}= 60 V$
For maximum voltage, resistance required, ${R}_{1}= 20\Omega$
Operating voltage, ${V}_{operate}= 75 V$
We know, according to Ohm’s law,
$V=IR$ …(1)
Where, V is the potential difference
I is the current flowing
R is the resistance
For maximum voltage, the equation. (1) can be written as,
${V}_{max}=I{R}_{1}$
Substituting values in above equation we get,
$60= I \times 20$
$\Rightarrow I= 3 A$
Now, to operate the lamp from 75 V supply, equation. (1) can be written as,
${V}_{operate}= I{R}_{2}$
$\Rightarrow 75= 3{R}_{2}$
$\Rightarrow {R}_{2}= 15 \Omega$
Therefore, the additional resistance required will be,
$R= {R}_{2}-{R}_{1}$
Substituting values in above equation we get,
$R= 20-15$
$\Rightarrow R=5\Omega$
Thus, the series resistance (in ohms) required to operate the lamp from a 75 V supply is $5\Omega$.
So, the correct answer is option D i.e. 5.
Note:
Students must remember that there are few limitations of Ohm’s law. Ohm’s law is not applicable for unilateral electrical elements like diodes and transistors as they allow the flow of current in only one direction. It is also not necessary that all conductors obey Ohm’s law. There are semiconductors such as silicon and germanium which do not obey Ohm’s law and these semiconductors are known as Non-ohmic conductors.
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