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Current Electricity — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsCurrent Electricity5 Marks
Question
  1. State Kirchhoff’s rules for an electric network. Using Kirchhoff’s rules, obtain the balance condition in terms of the resistances of four arms of Wheat sone bridge.
  2. In the meter bridge experimental set up, shown in the figure, the null point ‘D’ is obtained at a distance of 40 cm from end A ofthe meter bridge wire. If a resistance of 10 Ω is connected in series with $R_1$, null point is obtained at AD = 60 cm. Calculate the value of $R_1$ and $R_2.$
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Answer

  1. Kirchhoff’s Rule
    1. At any junction, the sum of the currents entering the junction is equal to the sum of the currents leaving the junction.
    2. The algebraic sum of the charges in potential around any closed loop involving resistors and cells in the loop is zero.
Condition of balance of a Wheatstone bridge: The circuit diagram of Wheatstone bridge is shown in fig.
P, Q, R and S are four resistance forming a closed bridge, called Wheatstone bridge. A battery is connected across A and C, while a galvanometer is connected between B and D. At balance, there is no current in galvanometer.
Derivation of Formula: Let the current given by battery in the balanced position be I. This current on reaching point A is divided into two parts $I_1$ and $I_2$ . As there is no current in galvanometer in balanced state, current in resistances P and Q is $I_1$ and in resistances R and S it is $I_2 .$​​​​​​​

Applying Kirchhoff’s I law at point A
$I - I_1 - I_2 = 0$ or $I = I_1 + I_2 - - - - - - - - (i)$
Applying Kirchhoff’s II law to closed mesh ABDA
$- I_1P + I_2R = 0$ or $I_1P = I_2 R - - - - - - - (ii)$
Applying Kirchhoff’s II law to mesh BCDB
$- I_1Q + I_2S = 0$ or $I_1Q = I_2S - - - - -- - (iii)$
Dividing equation (ii) by (iii), we get
$\frac{\text{I}_{1}\text{P}}{\text{I}_{1}\text{Q}} = \frac{\text{I}_{2}\text{R}}{\text{I}_{2}\text{S}}\text{ or }\frac{\text{P}}{\text{Q}} = \frac{\text{R}}{\text{S}}$ - - - - - - -(iv)
This is the condition of balance of Wheatstone bridge.
  1.  

For null point at D, balance length $\ell_{1}$ = 40 cm
So, $\frac{\text{R}_{1}}{\text{R}_{2}} = \frac{\text{AD}}{\text{DC}} =\frac{40}{(100- 40)} = \frac{2}{3}$ - - - -- - (i)
If resistance 10 $\Omega$ is connected in series of $R_1$, then balance length AD' > AD i.e. balance point shifts by length ‘y’ towards C i.e., AD = 60 cm.
$\frac{\text{R}_{1} + 10}{\text{R}_{2}} = \frac{\text{AD}"}{\text{D}'\text{C}} = \frac{60}{100-60} = \frac{3}{2}$
$\frac{\text{R}_{1}}{\text{R}_{2}} + \frac{10}{\text{R}_{2}} =\frac{3}{2}$- - - - - -- -- - -(ii)
From equations (1) and (2), we have
$\frac{2}{3} +\frac{10}{\text{R}_{2}} =\frac{3}{2}$
$\frac{10}{\text{R}_{2}} = \frac{3}{2} - \frac{2}{3} =\frac{9-4}{6} = \frac{5}{6}$
$\Rightarrow\text{R}_{2} = \frac{ 10\times6}{5} = 12\text{ ohm}$
From equation (1), we have
$\frac{\text{R}_{1}}{12} = \frac{2}{3}\Rightarrow\text{R}_{1} = \frac{12\times2}{3} =8\text{ ohm}.$

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