A resistor is fabricated by connecting two wires of the same material .The radii of the two wires are 1 mm and 3mm respectively and their lengths are 3 cm and 5 cm respectively. IF the two ends of the resistor are connected to the two terminals of a battery of emf 16 V and of negligible Internal resistance, what will be the potential difference between the two ends of the wire of shorter length?

Explanation

Resistance of wire, `R=rhol/A=rho.1/(pir^2)`
So for the two wires of same material,
`R_1/R_2=l_1/l_2timesr_2^2/r_1^2=3/5times(3/1)^2=27/5`
The two wires are connected in series, So, the ratio of the potential difference across their ends is
`V_1/V_2=(IR_1)/(IR_2)=R_1/R_2=27/5or,V_1/(V_1+V_2)=27/(27+5)=27/32`
or,`V_1=27/32(V_1+V_2)`
In the circuit there are only two resistances. So the electromotive force,
`E=V_1+V_2=15thereforeV_1=27/32times16=13.5V`
So the potential difference across the resistance of length 3 cm =`V_1`=13.5 V