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when a capacitor of value 200 micro farad charged to 200 V is discharge separately through resistance of 2 ohm & 8 ohm then the heat produced in joule will respectively be????a) 4&16 b) 16&4 c) 4& 8 d)4&4
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when a capacitor of value 200 micro farad charged to 200 V is discharg...
Answer should be D As energy stored in capacitor is 1/2 CV^2 which is equal to 41/2 x 200x 10^-6 x 200 x 200=4 JIt will not depend on the resistance as it is ask for separate condition Total energy stored in the capacitor will expelled in form of heat..... Therefore in both conditions 4 J heat is produced ... Hope that this should be the answer..
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when a capacitor of value 200 micro farad charged to 200 V is discharg...
We can use the formula for the discharge of a capacitor through a resistor:
V(t) = V0 * e^(-t/RC)
Where:
V(t) = voltage across the capacitor at time t
V0 = initial voltage across the capacitor (200 V in this case)
t = time elapsed since the discharge started
R = resistance of the circuit (2 ohms)
C = capacitance of the capacitor (200 micro farad = 0.0002 farad)
At t = 0 (the start of the discharge), V(t) = V0 = 200 V
At t = infinity (long after the discharge), V(t) = 0 V
We can find the time constant RC by multiplying R and C:
RC = 2 ohms * 0.0002 farad = 0.0004 seconds
Now, let's find the voltage across the capacitor at different times:
At t = 0.0004 seconds:
V(0.0004) = 200 * e^(-0.0004/0.0004) = 100 V
At t = 0.001 seconds:
V(0.001) = 200 * e^(-0.001/0.0004) = 36.8 V
At t = 0.01 seconds:
V(0.01) = 200 * e^(-0.01/0.0004) = 0.67 V
As we can see, the voltage across the capacitor decreases rapidly at the beginning of the discharge, but it takes a long time to fully discharge.
V(t) = V0 * e^(-t/RC)
Where:
V(t) = voltage across the capacitor at time t
V0 = initial voltage across the capacitor (200 V in this case)
t = time elapsed since the discharge started
R = resistance of the circuit (2 ohms)
C = capacitance of the capacitor (200 micro farad = 0.0002 farad)
At t = 0 (the start of the discharge), V(t) = V0 = 200 V
At t = infinity (long after the discharge), V(t) = 0 V
We can find the time constant RC by multiplying R and C:
RC = 2 ohms * 0.0002 farad = 0.0004 seconds
Now, let's find the voltage across the capacitor at different times:
At t = 0.0004 seconds:
V(0.0004) = 200 * e^(-0.0004/0.0004) = 100 V
At t = 0.001 seconds:
V(0.001) = 200 * e^(-0.001/0.0004) = 36.8 V
At t = 0.01 seconds:
V(0.01) = 200 * e^(-0.01/0.0004) = 0.67 V
As we can see, the voltage across the capacitor decreases rapidly at the beginning of the discharge, but it takes a long time to fully discharge.
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when a capacitor of value 200 micro farad charged to 200 V is discharge separately through resistance of 2 ohm & 8 ohm then the heat produced in joule will respectively be????a) 4&16 b) 16&4 c) 4& 8 d)4&4
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when a capacitor of value 200 micro farad charged to 200 V is discharge separately through resistance of 2 ohm & 8 ohm then the heat produced in joule will respectively be????a) 4&16 b) 16&4 c) 4& 8 d)4&4 for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about when a capacitor of value 200 micro farad charged to 200 V is discharge separately through resistance of 2 ohm & 8 ohm then the heat produced in joule will respectively be????a) 4&16 b) 16&4 c) 4& 8 d)4&4 covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for when a capacitor of value 200 micro farad charged to 200 V is discharge separately through resistance of 2 ohm & 8 ohm then the heat produced in joule will respectively be????a) 4&16 b) 16&4 c) 4& 8 d)4&4.
when a capacitor of value 200 micro farad charged to 200 V is discharge separately through resistance of 2 ohm & 8 ohm then the heat produced in joule will respectively be????a) 4&16 b) 16&4 c) 4& 8 d)4&4 for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about when a capacitor of value 200 micro farad charged to 200 V is discharge separately through resistance of 2 ohm & 8 ohm then the heat produced in joule will respectively be????a) 4&16 b) 16&4 c) 4& 8 d)4&4 covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for when a capacitor of value 200 micro farad charged to 200 V is discharge separately through resistance of 2 ohm & 8 ohm then the heat produced in joule will respectively be????a) 4&16 b) 16&4 c) 4& 8 d)4&4.
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