Two capacitors of capacitance 3 uF and 6 pF are connected in series to a 100 V
battery. Calculate the equivalent capacitance and the voltage across each
capacitor.
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Answer:
ANSWER
Given : C
1
=3μF C
2
=6μF C
3
=6μF
Equivalent capacitance for series combination
C
eq
1
=
C
1
1
+
C
2
1
+
C
3
1
C
eq
1
=
3
1
+
6
1
+
6
1
⟹C
eq
=1.5μF
Thus the charge flowing through the circuit q=C
eq
V
q=1.5μF×120 volt =180μC
Potential difference across 3μF, V
1
=
C
1
q
=
3μF
180μC
=60 volt
atifmalikpak07:
wrong answer
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