Two equal resistors R are connected in parallel, and a battery of 12 V is connected across this combination A dc current of 100 mA flows through the circuit as shown below: The value of R is
240 Ω
CONCEPT:
Resistance:
Resistances in parallel:

\(\frac{1}{R} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}\)

\(⇒ \frac{1}{R_{net}} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}\)
\(⇒ \frac{1}{R_{net}} = \frac{1}{{{R}}} + \frac{1}{{{R}}}=\frac{2}{R}\)
\(⇒ R_{net}=\frac{R}{2}\)
As we know, in parallel combination the potential difference across the resistors remains the same. Therefore, according to ohm's law
⇒ V = IRnet
\(⇒ R_{net}=\frac{R}{2}=\frac{V}{I}=\frac{12\times10^3}{100}=120 V\)
⇒ R = 240 Ω
NOTE :

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