A potentiometer wire has length $$4$$ m and resistance $$8$$ Ω. The resistance that must be connected in series with the wire and an accumulator of e.m.f. $$2V$$, so as to get a potential gradient $$1mV$$ per cm on the wire is

Question

A potentiometer wire has length $$4$$ m and resistance $$8$$ Ω. The resistance that must be connected in series with the wire and an accumulator of e.m.f. $$2V$$, so as to get a potential gradient $$1mV$$ per cm on the wire is

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Required potential gradient $$=1mVcm-1=110Vm-1$$ Length of potentiometer wire, $$l=4$$ mSo potential difference across potentiometer wire $$=110$$×$$4=0.4$$ V                 . . . . .(i) In the circuit, potential difference across $$8$$Ω$$=I$$×$$8=28+R$$×$$8$$                     . . . . (ii)Using$$ equation $$(i) and (ii), we get $$0.4=28+R$$×$$8410=168+R$$,$$8+R=40$$∴  $$R=32$$Ω

A potentiometer wire has length 4 m and resistance 8 $Ω$ . The resistance that must be connected in series with the wire and an accumulator of e.m.f. 2V, so as to get a potential gradient 1mV per cm on the wire is

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A potentiometer wire has length 4 m and resistance 8 Ω. The resistance that must be connected in series with the wire and an accumulator of e.m.f. 2V, so as to get a potential gradient 1mV per cm on the wire is

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A potentiometer wire has length 4 m and resistance 8 $Ω$ . The resistance that must be connected in series with the wire and an accumulator of e.m.f. 2V, so as to get a potential gradient 1mV per cm on the wire is