Current Electricity
7.0 Wheatstone bridge
7.0 Wheatstone bridge
This an arrangement consists of four resistance $P$, $Q$, $R$ and $S$ in which one resistance is unknown and rest are known.
Wheatstone bridge is shown in the figure. The bridge is said to be balanced when deflection in galvanometer is zero, So, ${I_g} = 0$
Balance condition, $$\frac{P}{Q} = \frac{R}{S}$$
Derivation of balance condition.
Applying kirchhoff's second law to loop $ABDA$, $${I_1}P + {I_g}G - {I_2}R = 0$$
Again, applying kirchhoff's second law to loop $BCDB$, $$\left( {{I_1} - {I_g}} \right)Q - \left( {{I_2} + {I_g}} \right)S - G{I_g} = 0$$
In balanced condition,${I_g} = 0$, thus$${I_1}P - {I_2}R = 0$$
or,$${I_1}P = {I_2}R$$
and,$${I_1}Q - {I_2}S = 0$$
or,$${I_1}Q = {I_2}S$$
On dividing above equations,we get$${\frac{P}{Q} = \frac{R}{S}}$$
Since the values of $P$, $Q$, and $R$ are known, the value of unknown resistor $S$ can be found.
Sensitivity of Wheatstone bridge.
A Wheatstone bridge is said to be sensitive if it shows large deflection in galvanometer for a small change of resistance in the resistance arm.
The bridge is most sensitive when all four resistance is of the same order.
Question 20. Calculate the resistance between $A$ and $B$ of the given network shown in the figure.
Solution: The circuit network can be redrawn as shown in the figure.
As shown, it is a Wheatstone bridge under balance condition.
Therefore, no current flows through resistance $10\Omega $.
Resistance of arm $ACB$, $$\begin{equation} \begin{aligned} {R_{ABC}} = 1 + 2 \\ {R_{ABC}} = 3\Omega \\\end{aligned} \end{equation} $$
And, resistance of arm $ADB$, $$\begin{equation} \begin{aligned} {R_{ABD}} = 2 + 4 \\ {R_{ABD}} = 6\Omega \\\end{aligned} \end{equation} $$
Net resistance between $A$ and $B$, $$\begin{equation} \begin{aligned} \Rightarrow \frac{1}{{{R_{AB}}}} = \frac{1}{{{R_{ABC}}}} + \frac{1}{{{R_{ABD}}}} \\ \Rightarrow \frac{1}{{{R_{AB}}}} = \frac{1}{3} + \frac{1}{6} \\ \Rightarrow {R_{AB}} = 2\Omega \\\end{aligned} \end{equation} $$
Question 21. A potential difference of $1V$ is applied between points $A$ and $B$ shown in the network in the figure. Each of resistances in the network equals $R=1\Omega $. Calculate
(i) Equivalent resistance between points $A$ and $B$.
(ii) Magnitudes of current flowing in the arms $AFDCB$ and $AFECB$
Solution: The equivalent network is
It is a balanced wheatstone bridge because, $$\frac{{1\Omega }}{{1\Omega }} = \frac{{1\Omega }}{{1\Omega }}$$
Hence the resistance in arm $DE$ is ineffective because no current passes through it.
Resistance of arm $FDC$, $$\begin{equation} \begin{aligned} {R_{ABC}} = 1 + 1 \\ {R_{ABC}} = 2\Omega \\\end{aligned} \end{equation} $$
And, resistance of arm $FEC$, $$\begin{equation} \begin{aligned} {R_{ABD}} = 1 + 1 \\ {R_{ABD}} = 2\Omega \\\end{aligned} \end{equation} $$
These two resistance are in parallel combination. So, equivalent resistance $(R_{eq})$, $$\begin{equation} \begin{aligned} \Rightarrow \frac{1}{{{R_{eq}}}} = \frac{1}{{{R_{FDC}}}} + \frac{1}{{{R_{FEC}}}} \\ \Rightarrow \frac{1}{{{R_{eq}}}} = \frac{1}{2} + \frac{1}{2} \\ \Rightarrow {R_{eq}} = 1\Omega \\\end{aligned} \end{equation} $$
(ii) Total current in the circuit, $$I = \frac{{1V}}{{1\Omega }} = 1A$$
Total current in the circuit = Current through arm $AFDCB$ + Current through arm $AFDEB$.
Also, Current through arm $AFDCB$ = Current through arm $AFDEB$
So, $${I_{AFDCB}} = {I_{AFDEB}} = 0.5A$$
Question 22. Find the value of unknown resistance $X$, in the circuit shown in the figure, if no current flows through the section $AO$. Also, calculate the current drawn by the circuit from the battery of $\xi = 6V$ and negligible internal resistance.
Solution: The equivalent circuit of given circuit diagram is
The given circuit is a balanced Wheatstone bridge as no current flows through the section $AO$.
Using principle of wheatstone bridge,$$\frac{2}{4} = \frac{3}{X}$$
Therefore, unknown resistance X is $$X = \frac{{3 \times 4}}{2} = 6\Omega $$
Total resistance along $BAC$ $\left( {2 + 4 = 6\Omega } \right)$ and $BOC$ $\left( {3+ 6 = 9\Omega } \right)$ form a parallel combination.
Thus, the effective resistance between $B$ and $C$ is$$R = \frac{{6 \times 9}}{{6 + 9}}=3.6$$
Total resistance in circuit = $3.6 + 2.4 = 6\Omega $
Total Current in the circuit , $$I = \frac{{6V}}{{6\Omega }} = 1A$$
Question 23. Find the effective resistance between points $P$ and $Q$ of electric circuit shown in the figure.
Solution: Using Principle of Wheatstone bridge, when the potential difference is applied between points $P$ and $Q$, no current will flow in arms $AB$ and $BC$.
The resistance in arms $AB$ and $BC$ are ineffective.
Equivalent circuit is,
Effective resistance between $P$ and $Q$, $$\frac{1}{R} = \frac{1}{{4R}} + \frac{1}{{4R}} + \frac{1}{{2r}} = \frac{{R + r}}{{2Rr}}$$
or, $$R = \frac{{2Rr}}{{R + r}}$$
