In the given figure: {V}_{1}= {V}, {V}_{2}=\alpha {V}, {R}_{1}=\beta {R}, {R}_{2}=\gamma {R}, where \alpha, \beta, and \gamma are positive real numbers. The value of current I is

Correct Answer is : (α−1)γ4β(β+γ)VR

\frac{(\alpha-1) \gamma}{4 \beta(\beta+\gamma)} { {V}}/{ {R}}

\frac{(\alpha-1)}{4 \beta} \frac{V}{R}

\frac{(\alpha-1) \beta}{2 \gamma(\beta+\gamma)} { {V}}/{ {R}}

\frac{(\alpha-1)(\beta+\gamma)}{2 \beta \gamma} { {V}}/{ {R}}

AP EAPCET 06-Jul-2022 Shift 1 Paper

Section: Physics

Question : 107 of 160

Marks: +1, -0

In the given figure: V1=V,V2=αV,R1=βR,R2=γR, where α,β, and γ are positive real numbers. The value of current I is

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