Use Kirchhoff’s rules to determine the value of the current I1 flowing in the circuit shown in the figure.

          

 

 

 

                                             OR

The figure shows experimental set up of a meter bridge. When the two unknown resistances X and Y are inserted, the null point D is obtained 40 cm from the end A. When a resistance of 10 Ω is connected in series with X, the null point shifts by 10 cm.

 

 

 

 

 

 

Find the position of the null point when the 10 Ω resistance is instead connected in series with resistance ‘Y’. Determine the values of the resistances X and Y.

$$\begin{gathered} \mathrm{Using} \mathrm{Kirchoff}\text{'}\mathrm{s} \mathrm{first} \mathrm{law} \mathrm{at} \mathrm{junction} \mathrm{E}, \mathrm{we} \mathrm{get} \\ {\mathrm{I}}_3={\mathrm{I}}_3+{\mathrm{I}}_2 \\ \mathrm{In} \mathrm{loop} \mathrm{ABCDA}, \mathrm{using} \mathrm{Kirchoff}\text{'}\mathrm{s} \mathrm{second} \mathrm{law}, \mathrm{we} \mathrm{get} \end{gathered}$$

$$\begin{gathered} 80-20{\mathrm{I}}_2+30 {\mathrm{I}}_1=0 \\ \Rightarrow 2 {\mathrm{I}}_2-3 {\mathrm{I}}_1=8 [+ \mathrm{by} 10]\dots \left(\mathrm{ii}\right) \\ \mathrm{In} \mathrm{loop} \mathrm{ABFEA}, \mathrm{we} \mathrm{get} \\ 80-20 {\mathrm{I}}_2+20-20 {\mathrm{I}}_3=0 \\ \Rightarrow {\mathrm{I}}_2+{\mathrm{I}}_3=5 [+ \mathrm{by} 20]\dots \left(\mathrm{iii}\right) \\ \mathrm{Putting} \mathrm{the} \mathrm{value} \mathrm{of} {\mathrm{I}}_3 \mathrm{into} \left(\mathrm{iii}\right), \mathrm{we} \mathrm{have} \\ {\mathrm{I}}_2+({\mathrm{I}}_1+{\mathrm{I}}_2)=5\Rightarrow 2 {\mathrm{I}}_2+{\mathrm{I}}_1=5 \dots \left(\mathrm{ii}\right) \\ \mathrm{Solving} \mathrm{equations} \left(\mathrm{ii}\right) \mathrm{and} \left(\mathrm{iv}\right), \mathrm{we} \mathrm{get} \\ {\mathrm{I}}_1=-\frac{3}{4}\mathrm{A}=-0.75 \mathrm{A} \end{gathered}$$

So (-) sign of current indicates that the direction of current is opposite to that as shown in the circuit diagram

                                  OR

$$\begin{gathered} \mathrm{With} \mathrm{X} \mathrm{and} \mathrm{Y}, \frac{\mathrm{X}}{\mathrm{Y}}=\frac{40}{60}=\frac{4}{6}\Rightarrow 6\mathrm{X}=4\mathrm{Y} \\ \mathrm{With} (\mathrm{x}+10) \mathrm{and} \mathrm{Y}, \\ \frac{\mathrm{X}+10}{\mathrm{Y}}=\frac{5}{5} \Rightarrow \mathrm{X}+10=\mathrm{Y} \\ \mathrm{Using} \mathrm{these} \mathrm{two} \mathrm{equations}, \mathrm{we} \mathrm{get} \frac{4\mathrm{Y}}{6}+10=\mathrm{Y} \\ \mathrm{i}.\mathrm{e}., \left(1-\frac{4}{6}\right)\mathrm{Y}=10 \mathrm{or} \mathrm{Y}=\frac{10\times 6}{2}=30 \mathrm{\Omega } \\ \therefore \mathrm{X}=\frac{4}{6}\times \mathrm{Y}=\frac{4}{6}\times 30=20 \mathrm{\Omega } \\ \mathrm{If} 10 \mathrm{\Omega } \mathrm{is} \mathrm{in} \mathrm{series} \mathrm{with} \mathrm{Y}, \mathrm{then} \frac{\mathrm{X}}{\mathrm{Y}+10}=\frac{\mathrm{l}}{100-\mathrm{l}} \\ \Rightarrow \frac{20}{40}=\frac{\mathrm{l}}{100-\mathrm{l}} \left[\therefore \mathrm{Y}+10=30+10=40\right] \\ \Rightarrow 2000-20\mathrm{l}=40\mathrm{l} \Rightarrow 60\mathrm{l}=2000 \\ \Rightarrow \mathrm{l}=\frac{200}{6}=33.3 \mathrm{cm}. \end{gathered}$$