(A)power supplied by the battery is 200 watt (B)current flowing in the circuit is 5A (C)potential difference across 4ohm resistance is equal to the potential difference across 6 ohm resistance (D)current in wire AB is zero
The conductor AB is connected in parallel to both the resistors (4 Ω and 6 Ω).
Hence it is short circuiting both the resistors. Therefore potential difference between both the resistors is same and it is equal to zero.
Now 2 Ω is the only resistance connected in the circuit with 20 V cell.
Therefore current flowing through the circuit (and through the wire AB) is 20/2= 10 A.
The power supplied by the battery is equal to heat dissipated by the resistor 2 Ω which is equal to (I2×R) = 102*2= 200 watt.
Therefore option a is correct.