In the circuit as shown in figure the

Q: In the circuit as shown in figure the

d $\frac{1}{{R'}} = \frac{1}{{10}} + \frac{1}{{10}} + \frac{1}{{20}}$$ \Rightarrow R' = \frac{{20}}{5} = 4\;\Omega $ Now using ohm’s law $i = \frac{{25}}{{R + R'}}$$ \Rightarrow 0.5 = \frac{{25}}{{R + 4}}$ $ \Rightarrow R + 4 = \frac{{25}}{{0.5}} = 50$ $ \Rightarrow R = 50 - 4 = 46\;\Omega $ Current through $20\,\Omega $ resistor $ = \frac{{0.5 \times 5}}{{20 + 5}} = \frac{{2.5}}{{25}} = 0.1\,A$ Potential difference across middle resistor $=$ Potential difference across $20\,\Omega $$ = 20 \times 0.1 = 2\,V$

SECTION - A [PHYSICS MCQ]

GSEB - English Medium

In the circuit as shown in figure the

AResistance $R = 46 \,\Omega$
BCurrent through $20\,\Omega$ resistance is $0.1 \,A$
CPotential difference across the middle resistance is $2\, V$
D
All option are correct

Medium

STD 11 Science
JEE
STD 12 - 3. current electricity
Physics

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