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Question: The effective capacitance between \(\mathrm{A}\) and \(\mathrm{B}\) will be ![](https://www.vedant...

The effective capacitance between A\mathrm{A} and B\mathrm{B} will be

A. 0.5F0.5F
B. 1.5F1.5F
C. 2F2F
D. 2.5F2.5F

Explanation

Solution

As a first step, the expression for efficient capacitance of parallel and series capacitor combinations could be recalled. Then, considering series and parallel combinations, you could approach the combination part by part. Now, and now, For each portion, apply the above expression and thus find the effective capacitance of the combination.

Formula used:
Effective capacitance,
For parallel,
Ceff=C1+C2C_{e f f}=C_{1}+C_{2}
For series,
1Ceff=1C1+1C2\dfrac{1}{C_{e f f}}=\dfrac{1}{C_{1}}+\dfrac{1}{C_{2}}

Complete answer:
In the question, we are asked to find the effective resistance across terminals A\mathrm{A} and B\mathrm{B}.

Before answering the question, let us recall the effective capacitance of parallel and series combinations.

When two capacitors are connected in series, then the effective capacitance is given by, 1Ceff=1C1+1C2\dfrac{1}{C_{e f f}}=\dfrac{1}{C_{1}}+\dfrac{1}{C_{2}}
When two capacitors are connected in parallel, then the effective capacitance is given by,
Ceff=C1+C2C_{e f f}=C_{1}+C_{2}

We see that in the given combination, capacitors XX and YY are connected in series. So the effective capacitance will be, CXY=CXCYCX+CYC_{X Y}=\dfrac{C_{X} C_{Y}}{C_{X}+C_{Y}}
CXY=2×22+2\Rightarrow C_{X Y}=\dfrac{2 \times 2}{2+2}
CXY=44=1μF\therefore C_{X Y}=\dfrac{4}{4}=1 \mu F

Now, we see that CXYC_{X Y} is connected parallel with capacitor Z\mathrm{Z}, their effective capacitance would be,
CXYZ=CXY+CZC_{X Y Z}=C_{X Y}+C_{Z}
CXYZ=1+1\Rightarrow C_{X Y Z}=1+1
CXYZ=2μF\therefore C_{X Y Z}=2 \mu F

The capacitance CC here is the capacitor CXYZC_{X Y Z}. We see that the capacitor CC is connected in series with the capacitor UU. Their
effective capacitance would be, CUC=UCU+CC_{U C}=\dfrac{U C}{U+C}
CUC=2×22+2\Rightarrow C_{U C}=\dfrac{2 \times 2}{2+2}
CUC=1μF\therefore C_{U C}=1 \mu F

Now we see that CUCC_{U C} is connected in parallel with the capacitor W\mathrm{W}.
Ceff=CUC+CWC_{e f f}=C_{U C}+C_{W}
Ceff=1+1\Rightarrow C_{e f f}=1+1
Ceff=2μF\therefore C_{e f f}=2 \mu F

Therefore, we found that the effective capacitance of the given combination to be 2μF2 \mu F.

Note:
Capacitance can be defined as the ratio of electric charge that is stored in a conductor to the potential difference across it. It could be otherwise defined as the ability of the body to hold an electric charge. The SI unit of capacitance is known to be Farad (F). The capacitance of the capacitor is dependent on its geometry.