
The cold resistance of a certain gas filled with tungsten lamp is 18.2Ohms and its hot resistance operating voltage of 110 V is 202Ohms, find the current
1) at the instant of switching on
2) The normal working current.
Answer
568.8k+ views
Hint: In this question we have been asked to calculate the current in the tungsten filament at the instant when it is switched and when it is working normally. We have been given the operating voltage as 110 V. Its starting resistance or cold resistance is given as 18.2 Ohms and hot or normal resistance is given as 202 Ohms. Therefore, we shall use Ohm's law to calculate the current.
Formula Used: \[I=\dfrac{V}{R}\]
Complete answer:
In this question we have been asked to calculate the current of the lamp at the instant when it is switched on.
From Ohm's law, we know,
\[I=\dfrac{V}{R}\]
a) Let \[{{I}_{1}}\] be the current when the lamp is switched on, the filament is now cold. Therefore, the resistance is 18.2 Ohms.
Therefore, we get
\[{{I}_{1}}=\dfrac{110}{18.2}\]
On solving,
\[{{I}_{1}}=6.04A\]
b) Now, the resistance of the lamp shall be taken as the hot resistance as the filament gets heated up after some time the lamp is on.
Therefore,
\[{{I}_{2}}=\dfrac{110}{202}\]
On solving,
\[{{I}_{2}}=0.54A\]
Therefore, the starting current is 6.04 A and the normal working current is 0.54 A.
Note:
From the above question we can say that the filament uses a larger amount of current to start than the normal working current. This happens because the filament takes time to heat up. Once the filament is heated the resistance increases and from Ohm's law, we know that current is inversely proportional to resistance. Therefore, the normal working current is usually less.
Formula Used: \[I=\dfrac{V}{R}\]
Complete answer:
In this question we have been asked to calculate the current of the lamp at the instant when it is switched on.
From Ohm's law, we know,
\[I=\dfrac{V}{R}\]
a) Let \[{{I}_{1}}\] be the current when the lamp is switched on, the filament is now cold. Therefore, the resistance is 18.2 Ohms.
Therefore, we get
\[{{I}_{1}}=\dfrac{110}{18.2}\]
On solving,
\[{{I}_{1}}=6.04A\]
b) Now, the resistance of the lamp shall be taken as the hot resistance as the filament gets heated up after some time the lamp is on.
Therefore,
\[{{I}_{2}}=\dfrac{110}{202}\]
On solving,
\[{{I}_{2}}=0.54A\]
Therefore, the starting current is 6.04 A and the normal working current is 0.54 A.
Note:
From the above question we can say that the filament uses a larger amount of current to start than the normal working current. This happens because the filament takes time to heat up. Once the filament is heated the resistance increases and from Ohm's law, we know that current is inversely proportional to resistance. Therefore, the normal working current is usually less.
Recently Updated Pages
Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

10 examples of friction in our daily life

