Here we have to calculate the total equivalent resistance and the expression for work
As per the question the two resistors are connected as parallel in a circuit.
Let [tex]R_{1} = 30[/tex]ohm and [tex]R_{2} =60[/tex]ohm
Let R is the total equivalent resistance
Hence [tex]\frac{1}{R} =\frac{1}{R_{1} } +\frac{1}{R_{2} }[/tex]
⇒ [tex]R=\frac{R_{1}*R_{2} }{R_{1}+R_{2} }[/tex]
= [tex]\frac{30*60}{30+60}[/tex]
=20 ohm
Here the potential difference is given as V=120 volt
Hence the current flowing through the wire is given as
[tex]I=\frac{V}{R}[/tex]
=[tex]\frac{120 volt}{20 ohm}[/tex]
=6 ampere[6A]
The work done is equal to the electric energy consumed by the circuit.
Let t is the time of consumption .Hence the work done is given as
[tex]W=I^{2}Rt[/tex]
=[tex]6^{2} *20*t[/tex] J
= 720t J
Taking various time of consumption we can get the work done
