Calculate the value of the resistance $R$ in the circuit shown in the figure so that the current in the circuit is $0.2A$. What would be the potential difference between points Band $E?$
CBSE OUTSIDE DELHI - SET 1 2012
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$R_{BCD} = 15 \Omega $
$ \frac{1}{\text{R}_{BE}} = \frac{1}{15} + \frac{1}{10} + \frac{1}{30} $
$ = \frac{2+ 3 + 1 }{30} $
$ \text{R}_{BE} = 5 \Omega $
$(15+5+R) \times 0.2 = 8-3= 5$
$(20+R) = 25$
$R = 5 \Omega $
$ \therefore\text{V}_{BE} = \text{IR}_{BE} $
$ = 0.2\times5 = 1.0 \text{V}.$
$ \frac{1}{\text{R}_{BE}} = \frac{1}{15} + \frac{1}{10} + \frac{1}{30} $
$ = \frac{2+ 3 + 1 }{30} $
$ \text{R}_{BE} = 5 \Omega $
$(15+5+R) \times 0.2 = 8-3= 5$
$(20+R) = 25$
$R = 5 \Omega $
$ \therefore\text{V}_{BE} = \text{IR}_{BE} $
$ = 0.2\times5 = 1.0 \text{V}.$
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