5 the following circuit improves the loading induced...

5. The following circuit improves the loading-induced nonlinearity in a potentiometer. $(1-\alpha)R_T$ $R_m$ $V_r$ $\alpha R_T$ $V_m$ $R_m$ (a) Express $v_m$ with $V_r$, $\alpha$, $k$. (Assume that $k = R_m/R_T$) (b) Find the relative error $\epsilon = (v_m - v_o)/v_o$ and plot $\epsilon$ against the change in $\alpha$. (c) When the relative error becomes zero?

Transcript

00:01 In this problem of dc circuits we have given a figure let me show you first.

00:06 So here is the figure and now in part a, say this is the figure for part a and we have to obtain the voltage v0 so we have to obtain this voltage v0 in this circuit.

00:21 So as you can see from here we can apply the current division so so we can say current in this section i0 would be say i0 is equal to say this is 4 plus 1 is equal to 5 kilo -oom and this is 5 kilo -oom so 5 kilo -oom and 5 kilo -oom are in parallel so this would be 5 divided with 5 plus 5 and this is multiplied with the current which is 2 millie -ampere so this is basically half of that so this would be 1 mili -amper and now we have to find the value of v0 so v0 is equal to i0 is equals to 1mm millie amp multiplied with 4 kilo ome is basically simply 4 volts so 4 volt is your right answer for part a and now part b so in part b we have to determine the voltage v0 v0 dash so this is here v0 dash or we can write it as we not this here measured when volt meter with 6 kilo ome internal resistance is connected so here this is having say resistance as 4 kilo ome and now here we have connected at this terminal this is a voltmeter is connected and now we have to find the value of v0 dash so first we would find the equivalent resistance of 4 om and 6 kilo om so this would be 4 kilo ome parallel with 6 kilo -oom would give you this is 4 kilo -oam and 6 kilo -oom so this is 4 kilo -oam parallel with 6 kilo -oom this is 4 kilo ome now this is equals to 4 multiplied with 6 divided with 10 so this is equal to 2 .4 kilo oom and now we can find i0 -d -dash so this is i not dash is equal to here this would be i not is i0 is equals to this would be say 5 say this is 5 and divided with 1 plus 2 .4 plus 5 and this is the full scale current so this would be 2 millie ampere now in the solvers so this is equal to 1 .19 milli ampere so we can find the value of v0 dash so this would be i0 dash multiplied with the resistance so this would be 2 .4 so this is 2 .4 so this would be 2 .4 kilo -oom since these are connected in parallel so register say voltage would be the same because in parallel combination we have the same voltage so this is multiplied with 1 .19 millie ampere so this is equals to 2...

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