Two capacitors A and B are connected in series across 100 volt supply ...
Problem: Two capacitors A and B are connected in series across 100 volt supply and it is observed that the potential difference across them 60 volt and 40 volt a capacitor of 2 micro Farad is now connected in parallel with A and potential difference across B Rises to 90 volt determine the capacitance of A and B?
Solution:
Step 1: Calculate the equivalent capacitance of capacitors A and B in series.
The potential difference across A and B is 60V and 40V respectively when they are connected in series. Therefore, using the voltage divider rule, we can find the equivalent capacitance of A and B in series.
Ceq = (C1 * C2) / (C1 + C2)
Where C1 and C2 are the capacitances of A and B respectively.
Substituting the given values, we get:
Ceq = (A * B) / (A + B)
= (60 * 40) / (60 + 40)
= 24 micro Farad
Step 2: Calculate the capacitance of capacitor A.
When a capacitor of 2 micro Farad is connected in parallel with capacitor A, the potential difference across B rises to 90V. Therefore, using the voltage divider rule again, we can find the capacitance of A.
Vb = (Ca / (Ca + Cp)) * Veq
Where Ca is the capacitance of A, Cp is the capacitance of the capacitor connected in parallel with A, Veq is the equivalent voltage across A and B (which is 100V) and Vb is the potential difference across B (which is 90V).
Substituting the given values, we get:
90 = (Ca / (Ca + 2)) * 100
Ca = 18 micro Farad
Step 3: Calculate the capacitance of capacitor B.
Now that we know the capacitance of A, we can find the capacitance of B using the equation we used in step 1.
Ceq = (Ca * Cb) / (Ca + Cb)
Substituting the given values, we get:
24 = (18 * Cb) / (18 + Cb)
Solving for Cb, we get:
Cb = 72 micro Farad
Therefore, the capacitance of capacitor A is 18 micro Farad and the capacitance of capacitor B is 72 micro Farad.
Two capacitors A and B are connected in series across 100 volt supply ...
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