Kirchhoffâ€™s law for capacitor:

(i) Junction law: It states that in any isolated system of capacitor the net charge is conserved. So, the incoming charge at any junction is equal to the outgoing charge from the junction.

At point P the charge incoming is taken as positive and charge outgoing is taken as negative. So, q_{1} â€“ q_{2} â€“ q_{3} = 0.

(ii) Loop law: The algebraic sum of voltage drop in a closed circuit of capacitor is zero. For the close loop ABCDA we get, V_{A} â€“ V_{B} =

V_{B} â€“ V_{C} =

V_{C} â€“ V_{D} =

So, V_{A} â€“ V_{D} = Â + Â + Â —— (i)

Again V_{A} â€“ V_{D} = Â ——- (ii) where Â is the emf of cell.

Â

From equation (i) and equation (ii) we get Â = Â + Â + Â or, Â + Â + Â – Â = 0.

Example: Find the potential difference between the points A and B using Kirchhoffâ€™s rule.Â

Let us consider q charge is entering at point B where q_{1} charge passes through E_{2} and (q â€“ q_{1}) charge passes through C_{3}. Using KVL for the loop PBASP we get, (V_{P} â€“ V_{B}) + (V_{B} â€“ V_{A}) + (V_{A} â€“ V_{S}) + (V_{S} â€“ V_{P}) = 0

Or, – E_{1} + Â + Â + 0 = 0

Or, qC_{1} â€“ q_{1}C_{1 }+ qC_{3} = E_{1}C_{1}C_{3}

Or, q(C_{1} + C_{3}) â€“ q_{1}C_{1} = E_{1}C_{1}C_{3} ——-(i)

Using KVL for the loop BQRAB we get (V_{B} â€“ V_{Q}) + (V_{Q} â€“ V_{R}) + (V_{R} â€“ V_{A}) + (V_{A} â€“ V_{B}) = 0

Or, – E_{2} + 0 + Â – Â = 0

Or, qC_{1} â€“ q_{1}C_{1 }+ qC_{3} = E_{1}C_{1}C_{3}

Or, q(C_{1} + C_{3}) â€“ q_{1}C_{1} = E_{1}C_{1}C_{3} ——-(i)

Using KVL for the loop BQRAB we get (V_{B} â€“ V_{Q}) + (V_{Q} â€“ V_{R}) + (V_{R} â€“ V_{A}) + (V_{A} â€“ V_{B}) = 0

Or, – E_{2} + 0 + Â – Â = 0

Or, q_{1}C_{3} â€“ qC_{2} + q_{1}C_{2} = E_{2}C_{2}C_{3}

Or, â€“ qC_{2} + q_{1}(C_{2} + C_{3}) = E_{2}C_{2}C_{3} ——-(ii)

Subtracting equation (i) from equation (ii) we get, q(C_{1} + C_{2} + C_{3}) – q_{1}(C_{1} + C_{2} + C_{3}) = E_{1}C_{1}C_{3} – E_{2}C_{2}C_{3}

Or, q – q_{1} = .

Therefore V_{A} â€“ V_{B} = – Â = .

Nodal analysis of capacitive circuit:

(i) At first identify different nodes and write their potential in the circuit.

(ii) Identify the isolated systems in the capacitive circuit and make equations using conservation of charge.

(iii) Solve the equations to get the values of potential at different points.

Find the potential difference between the points A and B using this concept:

Let us consider the potential at point B is E_{1} so potential at point P and S are 0. The potential at point Q is (E_{1} + E_{2}).

Let the potential at point A is x so potential at point R is (E_{1} + E_{2}).

Now consider the isolating system and taking charge entering the point A is equal the charge leaving point A.

So q_{1} + q_{3} = q_{2} or, C_{1}x + C_{3}(x â€“ E_{1}) = C_{2}(E_{1} + E_{2} â€“ x)

Or, C_{1}x + C_{3}x â€“ C_{3}E_{1} = C_{2}E_{1} + C_{2}E_{2} â€“ C_{2}x

Or, x =

So, the potential difference between the points A and B is

V_{A} â€“ V_{B} = x â€“ E_{1} = Â â€“ E_{1} = .

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