(a) Calculate the resistance of the wire using the graph.

(b) How many `176Omega` resistors in parallel are required to carry 5A on a 220 V line?
(c ) Define electric power. Derive relation between power, potential difference and resistance.

Explanation

(a) Resistance of wire = Slope of the graph
According of Ohm's law,
`V = IR or R=(V)/(I)`
`therefore R=(V_(2)-V_(1))/(I_(2)-I_(1))=(4-2)/(0.2-0.1)=(2xx10)/(1) = 20 Omega`
(b) Resistance, `R'=176Omega " No of resistors"=n, " Current," I=5A`
Potential difference, `V=220 " volts, Resultant resistance "=R`
According to Ohm's law,
`V=IR " " R=(V)/(I) =(220)/(5)=44Omega`
`(1)/(R)=((1)/(R')+(1)/(R')+...n " times")=(1+1+1+...n" times")/(R')`
`(1)/(R)=(n)/(R') rArr n=(R')/(R)=(176)/(44)=4`
Thus 4 resistors of 176 `Omega` in parallel combination are required to carry 5A on a 220 V line.
(c ) Electric power is defined as the electrical work done per unit -time.
Power`=("Work done")/("Time taken")rArr P=(W)/(t)`
The work done, W by current, I when it flows for time t under potential difference V is given by, `W= V xx I xx t` joules
`because P=(W)/(T)" " therefore P=(V xxI xx t)/(t) " " therefore P=V xx I`
But `I= (V)/(R)`
So, `P = (V xx V)/(R) rArr P=(V^(2))/(R)`
Electric power `=(("Potential difference")^(2))/("Resistance")`