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In the circuits described below, R is an unknown variable resistance. Three parallel branches connect top nodes B, C, D to a common bottom rail. Left branch: 2 Ohm resistor between nodes B and A. Middle branch: 4 Ohm resistor between C and F. Right branch: variable resistor R between D and E. Bottom connections (all EMF sources): A-F: 4V source; F-E: voltage source varies by circuit (P:6V, Q:8V, R_circ:4V, S:14/3 V, T:6V); C-F branch also has a 6V source in series. List-I contains statements about current directions and zero-current conditions. Match each statement in List-I with the correct circuit(s) from List-II. List-I: (I) Current through 4 Ohm resistor can be zero. (II) Current through 4 Ohm resistor can flow from F to C. (III) Current through 2 Ohm resistor can flow from B to A. (IV) Current through R is zero. List-II: (P) F-E source = 6V (Q) F-E source = 8V (R_circ) F-E source = 4V (S) F-E source = 14/3 V (T) F-E source = 6V (same as P) Note: Circuits P and T appear identical in the given data.

1. I -> P,Q,R,S,T; II -> Q; III -> P,Q,R,S,T; IV -> S
2. I -> Q; II -> P,Q,R,S,T; III -> P,Q,R,S,T; IV -> S
3. I -> P,Q,R; II -> P,Q,R,S,T; III -> T; IV -> P,Q,R,S,T
4. I -> T; II -> Q; III -> P,Q,R,S,T; IV -> P,Q,R,S,T

Correct answer: I -> Q; II -> P,Q,R,S,T; III -> P,Q,R,S,T; IV -> S

Solution

This is a complex circuit matching problem. Using Kirchhoff's voltage law systematically: with 4V between A-F, 6V in the middle branch, and the varying F-E source, we find that the 4 Ohm carries zero current only when the middle branch source exactly compensates (circuit Q with 8V on F-E allows this). Current through 2 Ohm can flow B to A in all circuits as R varies over full range. Current through R = 0 for a specific source value (circuit S with 14/3 V). The matching I->Q; II->all; III->all; IV->S corresponds to option (B).

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