The four arms of a Wheatstone bridge have resistance as shown in the figure. A galvanometer of $15$ resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of $10 V$ is maintained across AC.

$4.87 m A$
$4.87 μ A$
$2.44 μ A$
$2.44 m A$

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Solution

The correct option is A
$4.87 m A$

Step 1: Given data:
Resistance in the arm AB is = $100 Ω$
Resistance in the arm BC is = $10 Ω$
Resistance in the arm CD is = $5 Ω$
Resistance in the arm DA is = $60 Ω$
Resistance in the arm BD is = $15 R$
Step 2: Formula used:
$V = i R$ , where $V =$ potential difference, $i =$ current, $R =$ resistance.
KCL = $i_{1} + i_{2} + . . . at junction is zero$
Total current or charge entering a junction or node is exactly equal to the charge leaving the node as it has no other place to go except to leave, as no charge is lost within the node“.
Step 3: Calculation:
Let $V_{A}, V_{B}, V_{C}, V_{D}$ are voltages at point A, B, C and D respectively
On applying KCL for point B,
$\begin{array}{rcl} \frac{V_{B} - 10}{100} + \frac{V_{B} - V_{D}}{15} + \frac{V_{B} - 0}{10} & = & 0 \\ 3 V_{B} - 30 + 20 V_{B} - 20 V_{D} + 30 V_{B} & = & 0 \\ 53 V_{B} - 20 V_{D} - 30 & = & 0 \\ 53 V_{B} - 20 V_{D} & = & 30 . . . . . . . . . (1) \end{array}$
Similarly applying KCL for point D,
$\begin{array}{rcl} \frac{V_{D} - 10}{60} + \frac{V_{D} - V_{B}}{15} + \frac{V_{D} - 0}{5} & = & 0 \\ V_{D} - 10 + 4 V_{D} - 4 V_{B} + 12 V_{D} & = & 0 \\ - 4 V_{B} + 17 V_{D} & = & 10 \dots \dots \dots \dots (2) \end{array}$
After solving equation (1) & (2)
for this multiply equation (1 ) by $17$ and equation (2) by $20$
$\begin{array}{rcl} 901 V_{B} - 340 V_{D} & = & 510 \\ - 80 V_{B} + 340 V_{D} & = & 200 \end{array}$
After that add both equations-
$821 V_{B} = 710 V_{B} = 0.865 V o l t$
Now put the value of $V_{B} i n e q u a t i o n (1) -$ we get
$V_{D} = 0.792 v o l t V_{B} = 0.865 v o l t$
Step 4: Calculation of current through the galvanometer
Then the current through the galvanometer,
$\begin{array}{l} = \frac{V_{B} - V_{D}}{R} \end{array} \begin{array}{l} = \frac{0.865 - 0.792}{15} \end{array} = 4.87 m A$
Hence option A is the correct answer.

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