CBSE BoardEnglish MediumSTD 10ScienceElectricity3 Marks
Question
Find the value of current I in the circuit given below:
✓
Answer
$R_{A C}$ abd $R_{E D}$ are in parellel, so $\frac{1}{ R _{ p }^{\prime}}=\frac{1}{ R _{ AC }}+\frac{1}{ R _{ ED }}=\frac{1}{30}+\frac{1}{30}=\frac{1}{15}$ $\Rightarrow R _{ p }^{\prime}=15 \Omega$
Now $R _{ p }^{\prime}$ and $R _{ BC }$ are in series, so $R _{ s }= R _{ p }+ R _{ BC }=15+15=30 \Omega$
Again, $R _{ AB }$ and $R _{ p }^{\prime}$ are in parallel, so $\frac{1}{ R ^{\prime \prime}}=\frac{1}{ R _{ AB }}+\frac{1}{ R _{ s }^{\prime}}$
$=\frac{1}{15}+\frac{1}{30}=\frac{1}{10}$
$\therefore R^{\prime \prime}{ }_{p}=10 \Omega$
So, current flowing through the circuit is $I =\frac{ V }{ R " p }=\frac{3}{10}=0.3 A$
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