The four arms of a wheatstone bridge have resistances 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-

Applying KCL for point nbsp;B: V_B 10100+V_B V_D15+V_B 010=0 V_B 1020+V_B V_D3+V_B2=0 3V_B 30+20V_B 20V_D+30V_B=0 53V_B 20V_D=30 ....(1) Similarly apllyi ...

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Byju's Answer

Standard XII

Physics

Resistance and Resistors

The four arms...

Question

The four arms of a wheatstone bridge have resistances as shown in the figure. A galvanometer of $15 Ω$ resistance is connected across $B D$ . Calculate the current through the galvanometer when a potential difference of $10 V$ is maintained across $A C$ .

4.87 mA

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4.87 μ A

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2.44 μ A

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2.44 mA

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Solution

The correct option is A $4.87 mA$
Applying $K C L$ for point $B$ :

$\frac{V_{B} - 10}{100} + \frac{V_{B} - V_{D}}{15} + \frac{V_{B} - 0}{10} = 0$

$\frac{V_{B} - 10}{20} + \frac{V_{B} - V_{D}}{3} + \frac{V_{B}}{2} = 0$

$3 V_{B} - 30 + 20 V_{B} - 20 V_{D} + 30 V_{B} = 0$

$53 V_{B} - 20 V_{D} = 30$ ....(1)

Similarly apllying $K C L$ for point $D$ :

$\frac{V_{D} - 10}{60} + \frac{V_{D} - V_{B}}{15} + \frac{V_{D} - 0}{5} = 0$

$V_{D} - 10 + 4 V_{D} - 4 V_{B} + 12 V_{D} = 0$

$- 4 V_{B} + 17 V_{D} = 10$ ....(2)

After solving equation (1) & (2):
$V_{D} = 0.792$ volt
$V_{B} = 0.865$ volt

Then the current through the galvanometer, $i = \frac{V_{B} - V_{D}}{R}$

$i = \frac{0.865 - 0.792}{15}$

$i = 4.87 mA$

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

The four arms of a wheatstone bridge have resistances as shown in the figure. A galvanometer of $15 Ω$ resistance is connected across $B D$ . Calculate the current through the galvanometer when a potential difference of $10 V$ is maintained across $A C$ .