Find the equivalents capacitance between x and y
Answers
here is your answer
dude please refer to the attachment....
Let the capacitors between AC , CB , AD and BD be named as P , Q , R and S respectively
using the property of WHEATSTONE BRIDGE.........
In this circuit....
P = 5 uF
Q = 5 uF
R = 5 uF
S = 5 uF
●●》P/R = (5uF) / ( 5uF) = 1
●●》Q/S = ( 5uF) / (5uF) = 1
■ for a balanced wheatstone
bridge...
=> P/R = Q/S
in this given circuit,
P/R = Q/ S = 1
thus, it is a balanced wheatstone bridge.....
and hence, the capacitor of 10 uF can be removed or not included in the calculation of equivalent capacitance......
☆ for series combination of capacitance
1/ Cseries = 1/Ca + 1/Cb
☆ for parallel combination of capacitance
1/C parallel = Ca + Cb
The capacitor between AC and CB are in series combination.....
thus, Equivalent capacitance c =>
1/ Cc = 1/ Cac + 1/ Cbc
=> 1/ Cc = 1/ 5uF + 1/ 5uF
=> 1/ Cc = (1+1)/ 5uF
=> Cc = (5/2) uF
The capacitor between AD and DB are in series combination .....
Thus, the equivalent capacitance d =>
1/ Cd = 1 / Cad + 1 / Cdb
=> 1/ Cd = 1/ 5uF + 1/ 5uF
=> 1 / Cd = (1+1)/5uF
=> Cd = (5/2) uF
The capacitance C and D are in parallel combination......
therefore, the equivalent capacitance ab =>
Cab = Cc + Cd
=> Cab = 5uF/2 + 5uF/2
=> Cab = 10uF/2 = 5 uF
The capacitors of capacitance 5uF and 5 uF are in series combination .....
thus,,
the equivalent capacitance of whole
circuit { Ceq} =>
1/ C eq = 1/ C ab + 1/ 5 uF
=> 1/ C eq = 1/5 uF + 1/ 5uF
=> 1/ C eq = (1+1)/5 uF
=> C eq = (5/2)uF
ANSWER: -
equivalent capacitance =
2.5uF
here is your answer
dude please refer to the attachment....
Let the capacitors between AC , CB , AD and BD be named as P , Q , R and S respectively
using the property of WHEATSTONE BRIDGE.........
In this circuit....
P = 5 uF
Q = 5 uF
R = 5 uF
S = 5 uF
●●》P/R = (5uF) / ( 5uF) = 1
●●》Q/S = ( 5uF) / (5uF) = 1
■ for a balanced wheatstone
bridge...
=> P/R = Q/S
in this given circuit,
P/R = Q/ S = 1
thus, it is a balanced wheatstone bridge.....
and hence, the capacitor of 10 uF can be removed or not included in the calculation of equivalent capacitance......
☆ for series combination of capacitance
1/ Cseries = 1/Ca + 1/Cb
☆ for parallel combination of capacitance
1/C parallel = Ca + Cb
The capacitor between AC and CB are in series combination.....
thus, Equivalent capacitance c =>
1/ Cc = 1/ Cac + 1/ Cbc
=> 1/ Cc = 1/ 5uF + 1/ 5uF
=> 1/ Cc = (1+1)/ 5uF
=> Cc = (5/2) uF
The capacitor between AD and DB are in series combination .....
Thus, the equivalent capacitance d =>
1/ Cd = 1 / Cad + 1 / Cdb
=> 1/ Cd = 1/ 5uF + 1/ 5uF
=> 1 / Cd = (1+1)/5uF
=> Cd = (5/2) uF
The capacitance C and D are in parallel combination......
therefore, the equivalent capacitance ab =>
Cab = Cc + Cd
=> Cab = 5uF/2 + 5uF/2
=> Cab = 10uF/2 = 5 uF
The capacitors of capacitance 5uF and 5 uF are in series combination .....
thus,,
the equivalent capacitance of whole
circuit { Ceq} =>
1/ C eq = 1/ C ab + 1/ 5 uF
=> 1/ C eq = 1/5 uF + 1/ 5uF
=> 1/ C eq = (1+1)/5 uF
=> C eq = (5/2)uF
ANSWER: -
equivalent capacitance =
2.5uF