The circuit consists of two identical ohmic resistors with each having resistance R=3Ω and two identical nonlinear elements, whose voltage – current relationship looks like V=αI2, where α=2voltampere2. At which non-zero emf ε of the source the current through the galvanometer is equal to zero? Resistance of the source can be neglected

To utilize cross symmetry, take potentials as shown iR=ε2R should pass through non-ohmic element with V=ε2So, V=αI2⇒ε2=α(ε2R)2⇒ε=9 volt

The circuit consists of two identical ohmic resistors with each having resistance R=3Ω and two identical nonlinear elements, whose voltage – current relationship looks like V=αI2, where α=2voltampere2. At which non-zero emf ε of the source the current through the galvanometer is equal to zero? Resistance of the source can be neglected

To utilize cross symmetry, take potentials as shown iR=ε2R should pass through non-ohmic element with V=ε2So, V=αI2⇒ε2=α(ε2R)2⇒ε=9 volt

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The circuit consists of two identical ohmic resistors with each having resistance $R = 3 Ω$ and two identical nonlinear elements, whose voltage – current relationship looks like $V = α I^{2}, w h e r e α = 2 \frac{v o l t}{a m p e r e^{2}}$ . At which non-zero emf $ε$ of the source the current through the galvanometer is equal to zero? Resistance of the source can be neglected

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3 volt

6 volt

9 volt

18 volt

answer is C.

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Detailed Solution

To utilize cross symmetry, take potentials as shown

$i_{R} = \frac{ε}{2 R}$ should pass through non-ohmic element with $V = \frac{ε}{2}$

So, $V = α I^{2}$

$\Rightarrow \frac{ε}{2} = α {(\frac{ε}{2 R})}^{2} \Rightarrow ε = 9 v o l t$

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