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Asked in Re-NEET 2022 · Kirchhoff's laws in networks
R is connected across AC, which has resistance (R₀)/4; CB has (3R₀)/4 and is in series with that pair.
R_∥=((R₀/4)R)/(R₀/4+R)=(R₀R)/(R₀+4R).
Divider: V=V₀(R_∥)/(R_∥+3R₀/4)=V₀(R₀R)/(R₀R+3/4R₀(R₀+4R)).
V=V₀(4R)/(4R+3R₀+12R)=(4V₀R)/(3R₀+16R).
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