Learn steps to calculate determine equivalent capacitance for the given capacitor short circuit, practice series and parallel connection of capacitors.
\(Q_1=Q_2\)
A C1 and C2 are in series
B C1 and C3 are in parallel
C C2 and C3 are in series
D None
_{?}\(Q_1=C_1\,\Delta V_C\)
\(Q_2=C_2\,\Delta V_C\)
\(C_{eq}=\dfrac{Q}{\,\Delta V_C}\)
or, \(C_{eq}=\dfrac{Q_1+Q_2}{\,\Delta V_C}\)
or, \(C_{eq}=\dfrac{Q_1}{\,\Delta V_C}+\dfrac{Q_2}{\,\Delta V_C}\)
or, \(C_{eq}=C_1+C_2\)
Conclusion - If two or more capacitors are connected in parallel then their equivalent capacitance is given as
\(C_{eq}=C_1+C_2+C_3+...\)
A \(7\,\mu F\)
B \(9\,\mu F\)
C \(8\,\mu F\)
D \(4\,\mu F\)
\(\Delta V_1=\dfrac{Q}{C_1}\)
\(\Delta V_2=\dfrac{Q}{C_2}\)
\(\Delta V_C=\Delta V_1+\Delta V_2\)
\(\Delta V_C=\Delta V_1+\Delta V_2\).
\(\dfrac{1}{C_{eq}}=\dfrac{\Delta V_C}{Q}\)
or, \(\dfrac{1}{C_{eq}}=\dfrac{\Delta V_1}{Q}+\dfrac{\Delta V_2}{Q}\)
or, \(\dfrac{1}{C_{eq}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}\)
Conclusion - The equivalent capacitance of two or more capacitors connected in series, is given as
\(\dfrac{1}{C_{eq}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}+...\)
\(C_{eq}=\left (\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}+\dfrac{1}{C_4}+...\right)^{-1}\)
A \(8\,\mu F\)
B \(1\,\mu F\)
C \(2\,\mu F\)
D \(4\,\mu F\)
\(C_{eq_1}=C_1+C_2\) [C_{1} and C_{2} connected in parallel]
\(C_{eq_2}=C_3+C_4\) [C_{3} and C_{4} connected in parallel]
\(\dfrac{1}{C_{eq}}=\dfrac{1}{C_{eq_1}}+\dfrac{1}{C_{eq_2}}\) [\(C_{eq_1}\) and \(C_{eq_2}\)are in series]
or, \(\dfrac{1}{C_{eq}}=\dfrac{1}{(C_1+C_2)}+\dfrac{1}{(C_3+C_4)}\)
A \(1\,\mu F\)
B \(2\,\mu F\)
C \(4\,\mu F\)
D \(8\,\mu F\)
\(C_{eq_1}=C_1+C_2\) [C_{1} and C_{2} connected in parallel]
\(C_{eq_2}=C_3+C_4\) [C_{3} and C_{4} connected in parallel]
\(\dfrac{1}{C_{eq}}=\dfrac{1}{C_{eq_1}}+\dfrac{1}{C_{eq_2}}\) [\(C_{eq_1}\) and \(C_{eq_2}\)are in series]
or, \(\dfrac{1}{C_{eq}}=\dfrac{1}{(C_1+C_2)}+\dfrac{1}{(C_3+C_4)}\)
A \(2\,\mu F\)
B \(1\,\mu F\)
C \(1.5\,\mu F\)
D \(3\,\mu F\)
\(C_{eq}=C_1\)
Note - The shorted components are not damaged, they will function normally when short circuit is removed.
A \(6\,\mu F\)
B \(5\,\mu F\)
C \(1.5\,\mu F\)
D \(zero\)