Question

The voltage across a load is controlled by using the circuit diagram shown in figure. The resistance of the load and of the potential meter is R .The load is connected to the middle of the potentiometer . The input voltage is constant and equal to U. Determine the change in the voltage across the load if its resistance is doubled.

Answer 1

To solve potentiometer related questions we need to know its principle, which is the same as that of a wheatstone bridge. From the question we see that it doesn’t involve any direct application of the potentiometer, else it says that a load resistance R is connected to the middle of a potentiometer of resistance R ( in the middle means half of the resistance=$\dfrac{R}{2}$.

Complete step by step answer:
Now observing the diagram, we can see that their potentiometer along with the load resistance acts as a resistor. Now it forms two resistances, one is parallel between half of the potentiometer and the load resistor and the other system is the other half of the potentiometer. These two resistance systems are in series with each other.Thus resultant resistance:
$R' = \dfrac{R}{2} + (\dfrac{{R \times R/2}}{{R + R/2}})$ $= \dfrac{5}{6}R$

Now we have been given the value of input voltage , thus total current in the circuit is :
$I' = \dfrac{U}{{\dfrac{5}{6}R}} = \dfrac{6}{5}\dfrac{U}{R}$.
Thus the voltage across the load is:
${U_L} = U - I'\dfrac{R}{2} = \dfrac{2}{5}U$ .
Now it is said that the load resistance changes to 2R, hence the new resistance would be :
$R'' = \dfrac{R}{2} + (\dfrac{{2R \times R/2}}{{2R + R/2}})$ $= \dfrac{9}{{10}}R$

Thus new total current is:
$I'' = \dfrac{U}{{\dfrac{9}{{10}}R}} = \dfrac{{10}}{9}\dfrac{U}{R}$ .
Hence the value of the load voltage will be:
${U_{2L}} = U - I''\dfrac{R}{2} = \dfrac{4}{9}U$
Thus we can easily calculate the change as:
\therefore k = \dfrac{{{U_{2L}}}}{{{U_L}}}$$$$ = \dfrac{{10}}{9}.

Note: These types of questions are very important concept wise and are very useful in clearing the concepts of electricity. The steps on how to find the voltage across a load is to be remembered. Also the question should be read properly so that any information such as where the load is connected etc. can be found.