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In the circuit shown, all the ammeters are ideal. Match the following based on the circuit
Column-I Column-II
(A) If switch S is open, the ammeter(s) that read (s) less than 10 A (p) $A_{1} a n d A_{2}$
(B) If switch S is open, the ammeter(s) that read (s) equal current (q) $A_{4} a n d A_{5}$
(C) If switch S is closed, the ammeter (s) that read (s) more than 5 A (r) $A_{3} a n d A_{4}$
(D) If switch S is closed, the ammeter (s) that show (s) increase in the reading (s) $A_{6} a n d A_{5}$

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$(A) \to p, q, r, s; (B) \to p, q; (C) \to q, r; (D) \to q, r, s$

$(A) \to p, q, r, s; (B) \to q, r; (C) \to p, q; (D) \to q, r, s$

$(A) \to p, q; (B) \to p, s; (C) \to p, q; (D) \to p, q$

$(A) \to p, q; (B) \to q, s; (C) \to s, q; (D) \to p, q$

answer is D.

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Detailed Solution

When S is open:

$\begin{array}{l} I_{1} = \frac{15}{2} = 7.5 A \\ I_{2} = \frac{6}{2} = 3 A \\ L o o p A D C B A : \\ 15 - 20 + 6 + 10 = 2 I_{1} + 4 I_{3} + 2 I_{2} + 1 I_{3} \\ I_{3} = - 2 A \end{array}$

$\begin{array}{l} C u r r e n t i n A_{1} a n d A_{2} : \\ I_{3} - I_{1} = - 2 - 7.5 = - 9.5 A \\ I_{3} = - 2 A \\ C u r r e n t i n A_{4} a n d A_{5} : \\ I_{2} - I_{3} = 3 - (- 2) = 5 A \end{array}$

$A f t e r c l o \sin g S :$

$\begin{array}{l} I_{1} = \frac{15}{2} = 7.5 A \\ I_{2} = \frac{6}{2} = 3 A \end{array}$

$\begin{array}{l} I_{3} = \frac{10}{4} = 5 A \\ C u r r e n t i n A_{1} : \\ I_{1} + I_{4} = 7.5 - 7.5 = 12.5 A \end{array}$

$\begin{array}{l} C u r r e n t i n A_{2} : \\ I_{3} - I_{1} = 10 - 7.5 = 2.5 A \\ C u r r e n t i n A_{4} : \\ I_{2} - I_{3} = 3 - 10 = - 7 A \end{array}$

$\begin{array}{l} C u r r e n t i n A_{5 :} \\ I_{4} - I_{2} = 5 + 3 = 8 A \end{array}$

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