Current Electricity 2 Question 9

9. In the given circuit, the internal resistance of the $18 V$ cell is negligible. If $R _1=400 \Omega, R _3=100 \Omega$ and $R _4=500 \Omega$ and the reading of an ideal voltmeter across $R _4$ is $5 V$, then the value of $R _2$ will be

(2019 Main, 9 Jan II)

(a) $550 \Omega$

(b) $230 \Omega$

(c) $300 \Omega$

(d) $450 \Omega$

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Solution:

  1. According to question, the voltage across $R _4$ is 5 volt, then the current across it

According to Ohm’s law,

$$ \begin{gathered} \Rightarrow \quad V=I R \quad \Rightarrow \quad 5=I _1 \times R _4 \\ \Rightarrow \quad 5=I _1 \times 500 \\ \\ I _1=\frac{5}{500}=\frac{1}{100} A \end{gathered} $$

The potential difference across series combination of $R _3$ and $R _4$

$\Rightarrow \quad V _2=\left(R _3+R _4\right) I=600 \times \frac{1}{100}=6$ Volt

So, potential difference (across $R _1$ )

$$ V _1=18-6=12 V $$

Current through $R _1$ is,

$$ I=\frac{V _1}{R _1}=\frac{12}{400}=\frac{3}{100} A $$

So current through $R _2$ is,

$$ I _2=I-I _1=\frac{3}{100}-\frac{1}{100} A=\frac{2}{100} A $$

Now, from $V=I R$, we have,

$$ R _2=\frac{V _2}{I _2}=\frac{6}{(2 / 100)}=300 \Omega $$



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