SATHEE: Current Electricity 2 Question 26

Current Electricity 2 Question 26

26. An electrical circuit is shown in figure. Calculate the potential difference across the resistor of $400 Ω$ as will be measured by the voltmeter $V$ of resistance $400 Ω$ either by applying Kirchhoff’s rules or otherwise.

(1996, 5M)

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Solution:

The given circuit actually forms a balanced Wheatstone’s bridge (including the voltmeter) as shown below

$R_{V} = 400 Ω$

Here, we see that $\frac{P}{Q} = \frac{R}{S}$

Therefore, resistance between $A$ and $B$ can be ignored and equivalent simple circuit can be drawn as follows

The voltmeter will read the potential difference across resistance $Q$ .

$Currents i_{1} = i_{2} = \frac{10}{100 + 200} = \frac{1}{30} A$

$∴$ Potential difference across voltmeter

$= Q i_{1} = (200) \frac{1}{30} V = \frac{20}{3} V$

Therefore, reading of voltmeter will be $\frac{20}{3} V$ .

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Current Electricity 2 Question 26

26. An electrical circuit is shown in figure. Calculate the potential difference across the resistor of 400Ω as will be measured by the voltmeter V of resistance 400Ω either by applying Kirchhoff’s rules or otherwise.

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26. An electrical circuit is shown in figure. Calculate the potential difference across the resistor of $400 Ω$ as will be measured by the voltmeter $V$ of resistance $400 Ω$ either by applying Kirchhoff’s rules or otherwise.