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Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Omega maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emfof 1.02 V (for very moderate currents upto a few mA) gives a balance point at 67.3 cmlength of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 k Omegais put in series with it, which is shorted close to the balance point. The standard cell is then replaced bya cell of unknown emfe and the balance point found similarly, turns out to be at 82.3 cm length of the wire. (a) What is the value epsilon ? (b)What purpose does the high resistance of600 K Omega have ? (c) Is the balance point affected by this high resistance ? (d) Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0V instead of 2.0 V ? (e) Would the circuit work well for determining an extremely small emf, say of the order of a few mV (such as the typical emf of a thermocouple) ? If not, how will you modify the circuit ? (f) Can we use above circuit to measure very small emf of the order of mV. (For example, emf obtained in thermocouple) ? If not, then what change would you make ? |
Answer» <html><body><p></p>Solution :(a) We have , <br/> `(epsilon_(1))/(epsilon_(2)) = (l_(1))/(l_(2))` <br/> `therefore (1.02)/(epsilon) = (67.3)/(82.3)` <br/> `therefore epsilon = 1.02 xx (82.3)/(67.3)` <br/> `therefore epsilon = 1.247` V<br/> (b) Purpose of connecting high value resistance in <a href="https://interviewquestions.tuteehub.com/tag/secondary-638692" style="font-weight:bold;" target="_blank" title="Click to know more about SECONDARY">SECONDARY</a> circuit is to reduce the current through the galvanometer (and hence to save the galvanometer from the possible damage) when movable contact (jockey) is far from the balancing point. <br/> (<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>) No, because by connecting resistance in secondary circuit of a potentiometer,potential gradient existing on potentiometer wire does not changeand so balancing <a href="https://interviewquestions.tuteehub.com/tag/length-1071524" style="font-weight:bold;" target="_blank" title="Click to know more about LENGTH">LENGTH</a> also does not change. <br/> (d) Yes, because`epsilon_(1) = kl_(1) = ((epsilon rho )/(R + L rho + r) ) l_(1) =` <br/>constant . <br/>`rArr`As r change , `l_(1) ` has to change. <br/> `rArr` Balancing poin will be affected. <br/> (e) No, because in order to obtain balancing point on the potentiometer wire, emf of driver cell (connected in primary circuit) must be greater than emf of unknown cell (connected in secondary circuit).<br/> Note : See the explanation given below. <br/> For primary circuit of a potentimeter, <br/> `I = (epsilon)/(R + L rho + r) "" `... (1) <br/> (Where `rho`= resistance per unit length of apotentiometer wire ) <br/>According to formula of potential gradient, <br/> k = `(epsilon rho)/(R + L rho + r)= ((epsilon)/(R + L rho + r) ) rho = I rho `.... (2) <br/> If emf `epsilon_(1) ` is balaned by length `l_(1)` of potentiometer wire then, <br/> `epsilon_(1) = kl_(1)` <br/> `therefore (epsilon_(1))_(max) = K(l_(1))_(max)= ` kL = L `rho L "" ` .... (3) <br/>From equation (1),<br/> `epsilon = IR+ IL rho + Ir `<br/>` therefore epsilon = IR+(epsilon_(1))_(max)+ Ir `<br/> ` therefore epsilon -(epsilon_(1))_(max) = IR+ Ir ` <br/> `therefore epsilon - (epsilon_(1))_(max) gt `0 <br/> `therefore epsilon gt (epsilon_(1))_(max) ` <br/> above is the required result . <br/> (f) In order to measure very small emf accurately, we <a href="https://interviewquestions.tuteehub.com/tag/require-11720" style="font-weight:bold;" target="_blank" title="Click to know more about REQUIRE">REQUIRE</a> to have very small potential gradient on the potentiometer wire. Value of k can be reduced considerably by connecting proper series resistance with potentiometer wire. Also by <a href="https://interviewquestions.tuteehub.com/tag/increasing-1040633" style="font-weight:bold;" target="_blank" title="Click to know more about INCREASING">INCREASING</a> length of potentiometer wire, we can decrease the value of potential gradient.</body></html> | |