In the network shown, if power dissipated in the 6Ω resistor is 24 watt, reading of the ideal voltmeter V is

(1)If current through 6Ω resistor is i,Then i2 ×  6=24     ⇒    i=2A.The upper branch and lower branch of the circuit have the same resistance 2+4Ω   i.e.,  6ΩTherefore current through each branch is 1A.∴    VP+2×1−4×1=VQ  ⇒  VP−VQ=4-2=2  volt.