00:00Question is, a current of 5 Ampere through a battery is maintained for 30 seconds and
00:17in this time 600 Joule of chemical energy is transformed into electric energy.
00:26A. What is the EMF of the battery? B. How much electric power is available for heating and other uses?
00:37So first of all, we will form a data, is electric current I which is obtained is 5 Ampere and
00:49time is 30 seconds. Work which is equal to energy is obtained as 600 Joule. We have to find
01:08the emf of battery E in A and in B we will find the electric power. So, for solution work or energy
01:38is equal to V into I into T or EMF into current into time. So, EMF will be obtained as work upon current into time.
02:00Because, here current and time are multiplying with EMF on the other side of equal, it will divide.
02:07So, EMF will be equal to work or energy upon current into time. So, work or energy is obtained as 600 Joules divided by current which is 5 Ampere. So,
02:23time which is 30 seconds. So, after simplification we will get the value of EMF is 4 Volt. So, this is our first answer. Now, solution for B is
02:51power.
02:54Power is calculated by this formula. EMF into current or electric potential into current. So, power will be obtained as EMF is obtained as 4. So, 4 into current which is 5 Ampere. So, power will be obtained as 4 into 5. It will be
03:20power will be 20 Watt. So, this is our second answer.
03:27So, you will need to move forward. So, this is our second question. And, we need to follow the other questions.
03:30So, we need to move towards the current to current, which is our second to current to current and current.
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