Two rods are joined end to end, as shown. Both have a crosssectional area of $$0.01$$ $$cm2$$ . Each is $$1$$ meter long. One rod is of copper with a resistivity of $$1.7$$×$$10-6$$ $$ohm-centimeter$$, the other is of iron with a resistivity of $$10-5$$ $$ohm-centimeter$$. How much voltage is required to produce a current of $$1$$ ampere in the rods?

Question

Two rods are joined end to end, as shown. Both have a crosssectional area of $$0.01$$ $$cm2$$ . Each is $$1$$ meter long. One rod is of copper with a resistivity of $$1.7$$×$$10-6$$ $$ohm-centimeter$$, the other is of iron with a resistivity of $$10-5$$ $$ohm-centimeter$$. How much voltage is required to produce a current of $$1$$ ampere in the rods?

Infinity Learn · Accepted Answer

Here,  Length of each rod, $$l=1m$$  Area of $$cross-section$$ of each rod,  $$A=0.01$$ $$cm2=0.01$$×$$10-4$$ $$m2$$  Resistivity of copper rod,  ρ$$cu=1.7$$×$$10-6$$Ωcm       $$=1.7$$×$$10-6$$×$$10-2$$Ω$$m=1.7$$×$$10-8$$Ωm  Resistivity of iron rod,  ρ$$Fe=10-5$$Ωcm       $$=10-5$$×$$10-2$$Ω$$m=10-1$$Ωm ∴   Resistance of copper rod  $$RCu=$$ρCulA  and resistance of iron rod,  $$RFe=$$ρFelA As copper and iron rods are connected in series, therefore equivalent resistance is $$R=RCu+RFe=$$ρ$$CulA+$$ρ$$FelA=$$ρ$$Cu+$$ρFelA Voltage required to produce $$1$$ A current in the rods is  $$V=I$$ R $$=(1)RCu+RFe$$ $$=$$ρ$$Cu+$$ρFelA $$=1.7$$×$$10-8+10-710.01$$×$$10-4V$$ $$=10-7(0.17+1)106V=1.17$$×$$10-1$$ $$V=0.117$$ V

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